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PHY2048, Summary tasks, typical problems

Tasks units, speedometer, killing bats

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Keyword: ;
Submitted 08-24-2022, 18:36  By: coman

`10 "gallons"*("3.78Liters")/"1gallons" =37.8 "Liters"`

`d=v_"wave built into brain"*t_"measured"`

`"what's built in to the speedometer of car?"`

`"how many time the wheel rotates in one second "rarr " distance travelled in 1 sec"`

`"Circumference="2piR=pi*"diameter"_"wheel"`

tasks 2025 fall dimensional analysis, conversions

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`20 (m*(1(mi)/(1600m)))/(s*(1hr)/(3600s)) rarr (mi)/(hr)`
Task 1: would france give you a bigger kg pretending that's 1 kg in order to gain advantage when trading with US?
Task 2: if `T=f(m,k)` find explicit form !
Task 3: `"convert 400" ("ft"*1m/"3.(3)ft")/("hr"*((3600s)/"hr")) "into" m/s `

`1hr=60*60=3600sec`

Task 4: is `v=v_0+2a*x "dimensionally correct?" `

---

`v=d/t=(pi*"diameter"_"wheel"*"how many times the wheel rotated in 1 sec")/"1 sec"`

`"how to get rid of all bats in a cave?"`


`v_"wave"uarr=f(Tuarr)`
`"Task: use the bat analogy and answer : what happens if you replace the wheel on your car with a larger wheel ? "`

`"Larger wheel means greater diameter."`

`"Is the spedometer on your car going to underestimate or overestimate the speed of your car?"`

if diameter is 1 then

`v_"original wheel"=(pi*"diameter"_"original"*"no of rotations")/"time"`

lets say no of rotations is `10`


compare that with:

`v_"new wheel"=(pi*2*"diameter"_"original"*"no of rotations")/"time"`


`v_"original wheel"=(pi*1*10)/"1 sec"`

`v_"new wheel"=(pi*2*10)/"1 sec"`

``


Keyword: ;
Submitted 09-02-2022, 12:49  By: coman

Motion, properties used to characterizing motion, speed as slope of straight line

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`"slope"="rise"/"run"=(y_2-y_2)/(x_2-x_1)`

`"rise"/"run" rarr "slope"=tan(theta)="opposite"/"adjacent"`

`"properties used to characterizing motion"`

`v=d/t larr "slope"="rise"/"run"=(y_2-y_2)/(x_2-x_1)rarr 1.9 m/s`

`"slope"="rise"/"run"=(y_2-y_2)/(x_2-x_1)`

`v=d/t rarr m/s `

`a=(Deltav)/t=(v_f-v_i)/t rarr m/s^2`


Keyword: ;
Submitted 09-02-2022, 12:50  By: coman

`vec v=(dvec x)/(dt)`

`vec a=(dvecv)/(dt) rarr `

`v=intdv=inta*dt`

`vec a " has same sign as "vec Delta v=vec v_f)-vec v_i <0`


`"Area"="time"*"acc" rarr s*m/s^2=m/s rarr "area "velocity"`

``



Keyword: ;
Submitted 09-02-2022, 12:50  By: coman

`vec v=(dvec x)/(dt)`

`vec a=(dvecv)/(dt) rarr `

`v=intdv=inta*dt`

`vec a " has same sign as "vec Delta v=vec v_f)-vec v_i <0`


`"Area"="time"*"acc" rarr s*m/s^2=m/s rarr "area "velocity"`

Vectors: resolving into components:

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`tan(theta)=sin(theta)/(cos(theta)`
`sin(theta)="opp"/"hyp"`
`cos(theta)="adj"/"hyp"`

`tan(theta)=("opp"/"hyp")/("adj"/"hyp")="opp"/"adj"=y/x`

`|vec a|=x^2+y^2`
`theta=arctan(y/x)`
`x_1+x_2=x_"sum"`
`y_1+y_2=y_"sum"`

`vec x=xhati`

`vec y=yhatj`

`vec a*vec b=a*b*cos(theta)=`

`vec a timesvec b=a*b*sin(theta)=|((hati,hatj,hatk),(a_x,a_y,a_z),(b_x,b_y,b_z))|=hati(a_yb_z-a_zb_y)+hatj(a_zb_x-a_xb_z)+hatk(a_xb_y-a _yb_x)`

`10^2*10^4=100*10*10*10*10=10^(2+4)=10^6=1,000,000=100*10,000`

`10^4/10^2=10^(4-2)=10^2` `` ``



Keyword: ;
Submitted 09-02-2022, 13:05  By: coman

pressure, volume, temperature: state variables used to describing the state of a thermodynamic system

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`p, V, T rarr "state variables" rarr " used to describing the state of a thermodynamic system" `

`U=" internal energy"`

`Q= "heat anbsorbed or released by a system"`

Free fall dropping a rock, calculate speed after time `t`

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`"A rock is dropped from y=10 meters".`

`"Calculate its speed when " t=2 sec. `

`t=?`

`a="constant"=g=9.8 m/s^2 " each and every sec the speed increases by 9.8 "m/s`

`y=(t)=y_0+v_(0,y)*t+a_y*t^2/2`

`v_(o,"when dropped")=0 m/s`

`y-y_0=10 m=9.8*t^2/2`

`t_10=sqrt((2*10m)/(9.8m/s^2)) rarrsqrt(m/(m/s^2))=sqrt(m*s^2/m)=1.428 s`

`t_40=sqrt((2*40m)/(9.8m/s^2)) rarrsqrt(m/(m/s^2))=sqrt(m*s^2/m)=2.857 s`


`v(t)=v_(o)+a*t`

`v(t=2s)=0+9.8m/s^2*2 s rarr 19.6 m/s`

Calculate direction of a velocity vector!

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Keyword: ;
Submitted 09-09-2022, 22:20  By: coman

`d_x=a_x+b_x+c_x=46*sin(35.0)+46*sin(192)+46*sin(312) =23.4662m`

`d_x=a_x+b_x+c_x=46m*cos(35.0)+46m*cos(192)+46m*cos(312) =-17.364m`

`tan^-1([(17.364)/(23.462)]=180/pi*arctan((17.364)/(23.462))=36.504 rarr `

`theta_d=-36.504 " with respect to positive x axis"`

``

Chap 3 and Chap 4 PHY2048 Assignment Q 6 Hints:

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Keyword: ;
Submitted 09-10-2022, 09:19  By: coman

Chap 3 and Chap 4 PHY2048 Assignment Q 6 Hints:

`d_x=a_x+b_x+c_x=46*sin(35.0)+46*sin(192)+46*sin(312) =23.4662m`

`d_y=a_y+b_y+c_y=46m*cos(35.0)+46m*cos(192)+46m*cos(312) =-17.364m`

`tan^-1([(−17.364)/(23.462)]=180/pi*arctan((17.364)/(23.462))=36.504 rarr `

`theta_d=-36.504 " with respect to positive x axis"`

https://education.wiley.com/content/Halliday_Physics_10e/media/resource/video_solutions/03-017-EOC-prob-HRW.html


`180/pi*arctan((-17.364)/(23.462))=-36.504`

`46*sin(35.0)+46*sin(192)+46*sin(312)=-17.364083`

2 dimensional motion, equations of motion

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Keyword: ;
Submitted 09-14-2022, 12:49  By: coman


Eq of motion along x, motion `v="constant"`

`x=v_x*t`

`y=gt^2/2`

`v_x=v*cos(theta)`


`y=v_y*t=vsin(theta)*t`

`vsin(theta)*t=gt^2/2 rarr `

`vsin(theta)=gt/2 rarr `

`t=(2vsin(theta))/g`

Pluging time t in eq of motion for motion with constant g along y

`x=v_x*(2vsin(theta))/g)=v*cos(theta)*(2vsin(theta))/g`

`bar v=(d-v*cos(theta)*(2vsin(theta))/g)/((2vsin(theta))/g)`



Keyword: ;
Submitted 09-14-2022, 12:50  By: coman

gravitational acceleration

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`g=9.81 m/s^2`


`"velocity increase at a rate "of 9.8 m/s " each and very second sec"`


`y(t)=y_0+v_(0,y)*t+a_y*t^2/2rarr " motion with constant acceleration"`

`x(t)=x_0+v_(0,x)*t rarr " motion with constant speed"`

`sin(theta)="opp"/"hyp"=v_(0,y)/v`


`cos(theta)="adj"/"hyp"=v_(0,x)/v`


`"clear the gap: what does that mean ?"`

`" when x=2m "rarr y_f=|6m|`


`y(x=2m)=?=-6m`


`+ " up" rarr y_f-y_0=-6`

`g rarr -9.8 m/s^2`


`6=9.8/2*t^2`

`-6=-9.8/2*t^2`

2 dimensional motion setting up equations of motion!

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`y(t)=v_(0,y)*t+a_y*t^2/2rarr " motion with constant acceleration"`

`x(t)=v_(0,x)*t rarr " motion with constant speed"`

`(x(t))/v_(0,x)=t`

`y(x)=v_(0,y)*x/v_(0,x)+a_y*(x/v_(0,x))^2/2rarr "`


`tan(theta)="opp"/"adj"=v_(0,y)/v_(0,x)`

`y(x=2m)~~-6 m`


`sin(theta)=v_(0,y)/v`


`6m=2m*tan(40)+(9.8m/s^2)/2*(x/v_(0,x))^2`

`6m=2m*tan(40)+(9.8m/s^2)/2*((2m)/v_(0,x))^2`


`2/9.8*(6m-2m*tan(40))=((2m)/v_(0,x))^2`


`v_(0,x)^2=((2m)^2/(2/9.8*(6m-2m*tan(40))))`

`v_(0,x)=sqrt(((9.8m/s^2*2m)^2/(2(6m-2m*tan(40)))))`


`tan(theta)=v_(0,y)/v_(0,x) rarr `


`v_(o,y)=tan(40)*4.8297=4.052599 m/s`

`v_0=sqrt(v_o^2)=sqrt(v_(o,y)^2+v_(o,x))=sqrt(4.05259^2+4.8297^2)=6.30471 m/s`


`tan(theta) rarr theta=tan^-1(v_(o,y)/v_(o,x))`

`theta=180/pi*arctan(4.05259/4.8297)=39.99`

Equation of motion for motion with constant acceleration

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The entire physics :
Equation of motion for motion with constant acceleration along `y`, `-9.8 m/s^`

`y=y_0+v_(0,y)*t+g*t^2/2`

`" If + y is upwards" rarr g=-9.8 m/s^2`

`y-y_0=h=v_(0,y)*t-9.8*t^2/2`

`v_(0,y)=v_0*sin(theta)`

Equation of motion for motion with constant acceleration along `x`

`x=x_0+v_(0,x)*t`


`h=?`


`t=2/3 s`

`theta=56`

`tan(56)=v_y/v_x=`

Ballistic pendulum experiment

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`" Friction ignored:"`

`K rarr=mgh`

`v_0=?`

`mv^2/2 =mgh`

`v^2/2 =gh`

`"Friction included:"rarr "Kinetic energy of the projectile gets converted "`
`"into potential energy of the projectile+pendulum system + work done against frictional forces,"`
`"work done to lodge the projectile/bullet into the pendulum"`

`mv_"frictional"^2/2=mgh+W_"frictional"`


`v_"frictional"^2/2=gh+W_"frictional"/m`


``

Keyword: ;
Submitted 10-17-2022, 18:15  By: coman

`W=vecF *vec r=F*r*cos(theta)`


`W=int_(x_o)^(x_f)vec Fcdot d vec x`

`K=m*v^2/2`

`U_"gravitational"=mgh`

`U rarr K+W_"friction"`

Torque: definition, cross product

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`"An influence that tends to change the state of rotational motion"`


`vec tau=vec r times vec F`

`vec A=2hati+4hatj`

`vec B=5hati-10hatj`


`vec A times vec B=?vec B times vec A`

`vec A cdot vec B=?vec B cdot vec A`

`hati times hat i=1*1*sin(0)`


`hat j times hat k=-(hat k times hat j)`


`hati cdot hat i=1*1*cos(0)=1`

Boyle's Law

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`pdarr~1/(Vuarr)`


`pV=nRT rarr " if T = constant" rarr`


`pV="constant"rarr`


`p="constant"/V`

Lorentz Force upon a proton entering with velocity `v` into a magnetic field `B`

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`F_"Lorentz upon a proton"=?`

Proton enters an external mag field `B=2 T` with a `v= 4*10^5 m/s`

angle between


`vec v` and `vec B` `20^0`


`F_L=qvecv times vec B=+1.6*10^-19C*4*10^5 m/s*2T*sin(20)=4.377*10^-14N`

An electron moves with a speed of `8.0*10^6 m/s` along
the `+x` axis. It enters a region where there is a magnetic
field of `2 T`, directed at an angle of `60^o` to the `+x` axis
and lying in the `xy` plane.

`F_L=qvecv times vec B=qvBsin(theta)`


`vec a=(vec F_"Lorentz")/m_"electron"=? larr F_"Lorentz"=m*a`

`vec a=(-1.6*10^-19C*8.0*10^6 m/s*2T*sin(60))/(9*10^-31 kg)rarr m/s^2`

`T=N/(C*m/s)=N/C*s/m`


`(C*m/s*N/C*s/m)/(kg)=??=m/s^2`


`N=kg*m/s^2`

---

Centrifugal force experiment in class, theory, equations:

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---

`F_c=m*a_c=m*v^2/Rrarr N`

5 points   `F_"c at equator"=m_"student"*((2*pi*r_"earth")/(24"hr"*3600s/"hr"))^2/r_"earth"=?`

`r_"earth"=6400*10^3m`

5 points   `F_"c at north pole"=?`

10 points   `F_"1, centrifugal"=m_"1, spinning"*(d/t)^2/r`

`F_"1, centrifugal"=m_"1, spinning"*((2pi*r)/t)^2/r`

`F_"1, centrifugal"=m_"1, spinning"*((4pi^2*r)/t^2)`

10 points   `F_"1,balancing"=m_"1,hanging"*g`

10 points   `F_"2, centrifugal"=m_"2, spinning"*(d/t)^2/r`

`F_"2, centrifugal"=m_"2, spinning"*((4pi^2*r)/t^2)`

10 points   `F_"2,balancing"=m_"2,hanging"*g`

5 points   `% "error"=|M-A|/A*100 rarr %`

5 points   `"Objective"=?`

---

Angular momentum vector as a cross product

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`vec L=vec r times vec p`

`vec F=(dvecp)/(dt)`


`vec tau=(dvecL)/(dt)rarr vec tau_"ext"=0 rarr Deltavec L=0 rarr L_(f)=L_i`


`L=Idarr*omegauarr="constant"`


`m rarr I`


`F rarr tau`


`p rarr L`

`v rarr omega`


`a rarr alpha`


`F=ma`


`tau=I*alpha`


`p=mv`

`L=Iomega`

`omega^2=omega_i^2+2alpha*Deltatheta`

`(d(mv))/dt=m(dv)/(dt)+v(dm)/(dt)`


`m(v(t))`

Moment of inertia about a point/axis

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`I=intr^2dm=intr^2rho*dxdydz`

Find the center of mass of the 3 rods system

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`"top rod has a mass of 4 m, "`

`"Left right rods mass =m"`

`"uniform density of all 3 rods"`

`"rod 1, top"=(x_1,y_1)=(L/2,o)`


`"rod 2, left"=(x_2,y_2)=(0,-L/2)`


`"rod 3, right"=(x_3,y_3)=(L,-L/2)`


`vec r_(CM)=(x_(CM),y_(CM))`


`x_(CM)=(sum_(i=1)^3(m_ix_i))/(sum_(i=1)^3(m_i))=(m_1*x_1+m_2*x_2+m_3*x_3)/(m_1+m_2+m_3)`


`x_(CM)=(4m*L/2+m*0+m*L)/(4m+m+m)=(4m*L/2+2/2m*L)/(6m)=m(2L+L)/(6m)=3L/6=1/2L`

`y_(CM)=(sum_(i=1)^3(m_iy_i))/(sum_(i=1)^3(m_i))=(m_1*y_1+m_2*y_2+m_3*y_3)/(m_1+m_2+m_3)`

Elastic restoring force when compressing/extending a spring

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`F=-kx`


`k=F/x rarr N/m`


`k/m rarr (N/m)/(kg)=(kg*m/s^2)/(m*kg)=1/s^2`


`f rarr 1/s=Hzs^-1`


`x(t)=A*cos(omegat)`

`x(t)=A*sin(omegat)`

`omegatrarr 1/s*s`

What are the units for square root of g over Length?

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`"Task:" sqrt(g/L)=?1/s`


`sqrt(k/m)=1/s`

Simple harmonic motion, position, velocity acceleration

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`x(t)=A*sin(omegat)`

`v=(dx)/(dt)=omegaA(cos(omegat))`

`a=(dv)/(dt)=d/(dt)(omegaA(cos(omegat)))=-omega^2(sin(omegat))`


`omega=(d theta)/(dt)rarr "rad"/s`


`alpha=(domega)/(dt) rarr "rad"/s^2`


`omega=(2pi)/T`


`T=(2pi)/omega=`

`d/dt(omegat)=omega`


`x(t)=A*sin(omegat)`


`v_"max"=omegaAcos(omegat)`

`omega=(2pi)/T=2pif`

`a_"max"=omega^2A=(2pif)^2*A=(2pi*3)^2*7/100`


`y=f(x,t)`

`(del^2y)/(delx^2)=k(del^2y)/(delt^2)`


`y^2+2x-4=0`


`((delf)/(dely))_(|x="constant")=2y`


`((delf)/(delx))_(|y="constant")=2`


`E~(4A)^2`

rotational equilibrium

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`"rotational equilibrium "`

`sum_(i=1)^ntau_i=0`

`t_"CC" rarr +`

`t_"CW" rarr -`

`vec tau=vec r timesvec F`


`t_"CW"=-m_"diver"*g*l_3 larr " weight of the diver" rarr `

`t_"CC"=+F_1*l_1+F_2*l_2=F_1*2m+F_2*1m`

`F_1*2m+F_2*1m-m_"diver"*g*l_3 =0`


`l_3=? "such that when the diver moves to that point the force exerted by support 2 =0"`


`F_1*2m+0*1m-m_"diver"*g*l_3 =0 rarr `

`l_3=? `


`F_1*2m-m_"diver"*g*l_3 =0`


`F_1*2m=m_"diver"*g*l_3 `

`(F_1*2m)/(m_"diver"*g)=l_3 `

Mass moving down an incline with constant speed

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Hint For Question 14:

`"Frictional force, " F_(f) " is cancelled out by the " x" component of the weight", W_x:`

`F_F=W_x`

Frictional force is proportional with the normal, `N=mgcos(theta) rarr`

`mu_s*mgcos(theta)=mgsin(theta) rarr`

`mu_s=tan(theta)`

Doppler effect

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`f_"observer"=f_"source"*(v_"sound"+-v_"observer")/((v_"sound"+-v_"source"))`

`
"towards each other"`


`f_"observer" >f_"source"`


`(v_"sound"+-v_"observer")/((v_"sound"+-v_"source")) >1`


`(v_"sound"-v_"source")`


`f_"observer" =f_"source"(v_"sound"+v_"observer")/((v_"sound"-v_"source"))`


`c=3*23*10^8 m/s`

`v_"sound"=340 m/s`

Buoyant force

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`"Part I" `

`F_"buoyant from vol of fluid displaced"=W_"fluid displaced"`


`W_"fluid displaced"=m*g`

`W_"fluid displaced"=rho_"fluid"*V_"object"*g`


`rho=m/V rarr (kg)/m^3 `


`"Part II" `

`F_"buoyant, from weight diff"=W_"air"-W_"fluid"=98N-28.21 larr "accepted=A"`


`% "error"|M-A|/A*100 rarr %`


`uarrF_"buoyant"=uarrrho_"fluid"*V_"object"*g`

determining the volume thus density of an irregularly shaped object by completely submerging the object in water

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`"Part III, determining the density of an irregularly shaped object by completely submerging the object in water"`

`rho_"object"="mass"/"Volume" rarr (kg)/m^3`

Identify the material the object is made of by looking up a density table.

`"Part IV, determining the specific density, ratio of the density of the object divided by density of the fluid:"`


`"specific density"=rho_"object"/rho_"fluid" rarr`

`"calculate the percentage of the object above fluid"`

Example:

`rho_"ice"=900 (kg)/m^3`

`rho_"freshwater"=1,000 (kg)/m^3`

`"specific density"=rho_"ice"/rho_"freshwater"=(900 (kg)/m^3)/(1,000 (kg)/m^3)=0.9 rarr`

`90% " of an iceberg is under water"`

Scuba divers advise, pressure is proportional with depth

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Anyone who scuba dives is advised not to fly within the next 24 h because the air mixture for diving can introduce nitrogen to the bloodstream. Without allowing the nitrogen to come out of solution slowly, any sudden air-pressure reduction (such as during airplane ascent) can result in the nitrogen forming bubbles in the blood, creating the bends, which can be painful and even fatal. Military special operation forces are especially at risk. calculate the magnitude of the change in pressure on such a special-op soldier who must scuba dive at a depth of `y_"air"=18 m` in seawater one day and parachute at an altitude of `y_"air"=7.7 km` the next day? Assume that the average air density within the altitude range is `rho_"air"=0.85 (kg)/m^3`, `rho_"seawater"=1024(kg)/m^3`.

Since it asks for the change in pressure:



` p_"air"=p_"seawater"+rho_"seawater"*g*y_"seawater"−rho_"air"*g*y_"air"rarr`


` p_"air"-p_"seawater"=rhog(y_"seawater"−y_"air")rarr`

`Delta p`

where `p_"air"` is the pressure in air

Simple harmonic motion equation of motion;
Hooke's Law experiment

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Simple harmonic motion equation of motion:

`y_"SHM"=x_"max"*sin(omegat)`


`omega=(Deltatheta)/(Deltat)`


`Deltatheta=omega*t`

Traveling transverse wave, equation of motion:

`y(x,t)=A*sin(kx-omegat)`


`k="wave number"=(2pi)/lambda rarr 1/m`


`x rarr m`


`kxrarr 1/m*mrarr " unitless"`



`e^(-lambda*t)`


`omegat rarr "rad"/"sec"*"sec"`


`"pirad"=180^0`

`omega=(2pi)/T=2pi*nu`


`lambda=v/f)=(10 m/s)/(60 Hz)rarr m/s*s/1=m`


`k=(2pi)/lambda=(2pi)/(0.16666m)`


`x=Asin(omegat)`


`v_"max"=Aomega`

`v=lambda*f lambda=v/f`


`omega=(2pi)*f`

`k=(2pi)/lambda`


`v_"max"=omegaA`



`F=-kx`


`k=F/x=
(m*g)/x larr "slope"=(y_2-y_1)/(x_2-x_1)`

Travelling wave & its variable

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Travelling wave & its variable

https://www.geogebra.org/m/mEweX3HF#material/ESNP9PSY

Traveling wave equation

https://www.geogebra.org/m/YKuG3zNZ

Longitudinal and transverse waves

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Longitudinal and transverse waves

https://www.geogebra.org/m/auyft2pd


Keyword: ;
Submitted 08-21-2023, 11:59  By: coman

`v_"built in"=d/t_"measured" larr 70 "mi"/"hr"`

`d=v_"built in"*t_"measured" `

Electric Force, sodium chloride molecule

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`Na^+Cl^-`



`F_"el"=8.99*10^9*(q_1*q_2)/r^2`



`F_"g"=6.*10^-11*(m_1*m_2)/r^2`

`vectau=(dvecL)/(dt)`


`6.*10^-11*(m_1*m_2)/x^2=(d^2x)/(dt^2)`


`6.*10^-11*(m_1*m_2)/y^2=(d^2y)/(dt^2)`

`6.*10^-11*(m_1*m_2)/z^2=(d^2z)/(dt^2)`

`6.*10^-11*(m_1*m_3)/x^2=(d^2x)/(dt^2)`


`6.*10^-11*(m_1*m_3)/y^2=(d^2y)/(dt^2)`

`6.*10^-11*(m_1*m_3)/z^2=(d^2z)/(dt^2)`

`6.*10^-11*(m_2*m_2)/x^2=(d^2x)/(dt^2)`


`6.*10^-11*(m_2*m_2)/y^2=(d^2y)/(dt^2)`

`6.*10^-11*(m_2*m_2)/z^2=(d^2z)/(dt^2)`

Partial derivatives

Top of page


`f(x,y)=4x^2*y`

`(delf)/(delx)=(df)/(dx)=`


`|PsiPsi^**|`

`vectau=(dvecL)/(dt)`



Keyword: ;
Submitted 08-23-2023, 13:27  By: coman

`v_"built in"=d/t_"measured" larr 70 "mi"/"hr"`

`d=v_"built in"*t_"measured" `


`Na^+Cl^-`



`F_"el"=8.99*10^9*(q_1*q_2)/r^2`



`F_"g"=6.*10^-11*(m_1*m_2)/r^2`

`vectau=(dvecL)/(dt)`



`x(t)=x_0+v_o*t+a/2*t^2`


`m=m+m/s*s+m/s^2*s^2`

`6.*10^-11*(m_1*m_2)/x^2=(d^2x)/(dt^2)`


`6.*10^-11*(m_1*m_2)/y^2=(d^2y)/(dt^2)`

`6.*10^-11*(m_1*m_2)/z^2=(d^2z)/(dt^2)`

`6.*10^-11*(m_1*m_3)/x^2=(d^2x)/(dt^2)`


`6.*10^-11*(m_1*m_3)/y^2=(d^2y)/(dt^2)`

`6.*10^-11*(m_1*m_3)/z^2=(d^2z)/(dt^2)`

`6.*10^-11*(m_2*m_2)/x^2=(d^2x)/(dt^2)`


`6.*10^-11*(m_2*m_2)/y^2=(d^2y)/(dt^2)`

`6.*10^-11*(m_2*m_2)/z^2=(d^2z)/(dt^2)`

Dimensional anlysis principle, Finding explicit form of a physical law

Top of page


`x(t)=x_0+v_o*t+a/2*t^2`


`m=m+m/s*s+m/s^2*s^2`


`"Finding explicit form of a physical law"`


`T="period"`


`T(L,g)=??"explicti form"=??=L^3*sqrt(g)??`


`"known"`

`L rarr m`


`t rarrs`

`grarr m/s^2`

`s=m^x*(m/s^2)^y`


`T=L^x*g^y`


`x=?, y=?`

`m^0*s=m^x*(m/s^2)^y`

`m^0*s=m^(x+y)*(1/s^2)^y`


`m^0*s^1=m^(x+y)*s^(-2y)`

`0=x+y`

`1=-2y`


`y=-1/2`

`x=1/2`

`T=L^(1/2)*g^(-1/2)`



`T=sqrt(L/g)`

Unit vectors, calculate direction of a vector

Top of page

`vec r=(x,y,z)`

`vec tau=vec r times vec F=|(hat i, hat j, hatk),(x,y,z),(F_x,F_y,F_z)|`


`hat i=|1|`

`cos (theta)="adj"/"hyp"=a_x/|vec a|=cos(21.8)=a_x/26.9`


`a_x=26.9*cos(21,8)=25.11`

`a_y larr sin(theta)=a_y/|vec a|`


`a_y=26.9*sin(21,8)= 9.64`


`"Given only x and y components calculate theta"`


`tan(theta)=a_y/a_x rarr`


`theta=arctan(a_y/a_x)=tan^-1(a_y/a_x)`

`vecc=vec a+vec b`


`|vecc|=?=sqrt(c_x+c_y`


`c_x=a_x+b_x=|a|*cos(theta_1)+|b|*cos(theta_2)`

`c_y=a_y+b_b=|a|*sin(theta_1)+|b|*sin(theta_2)`


`theta_c=arctan(c_y/c_x)`

dot product using unit vectors

Top of page

`vec r=(x,y,z)`

`vec tau=vec r times vec F=|(hat i, hat j, hatk),(x,y,z),(F_x,F_y,F_z)|=hati(yF_z-zF_y)+hatj(zF_x-xF_z)+hatk(xF_y-yF_x)=|r||F|sin(theta)`


`hat i=|1|`


`vec acdot vec b=?=|a|*|b|*cos(theta)`

Task vector sum=?

Top of page

`a_x=10, a_y=20, theta_a=30`

`b_x=5, b_y=12, theta_b=110`

`Task 1: vec c=?=vec a+vec b rarr " both magnitude nd direction"`

Task 2 calculate vector sum, magnitude and direction

Top of page


`Task 2: vec c=?=vec a+vec b rarr " both magnitude and direction"`

`a_x=10, a_y=20, theta_a=30`

`b_x=15, b_y=12, theta_b=110`

`c_x=10-15=-5`

`c_y=20+12=32`

`|c|=sqrt((-5)^2+32^2`

`theta_c=phi=arctan(c_y/c_x)`

``


Keyword: ;
Submitted 09-06-2023, 09:06  By: coman

`v=d/t`


`a=(Deltav)/t=(v_f-v_0)/(Deltat) rarr vec a=(d^2vec r)/(dt^2)`


`g=9.81 m/s^2`



`F_f(v)`


`int_0^(t_f)a*dt=int_(v_i)^(v_f)(dvecv) rarr `


`vec F=m(dv)/(dt)`


eq motion for motion with cnst acc

`y(t)=y_0+v_0*t+g*t^2/2`


`y(t)=g*t^2/2`

`t=sqrt(2*y/g)`


`%error=|M-A|/A*100=|9.1-9.81|/9.81*100 rarr %`



`g="slope"=9.1 m/s^2`

Speed a scalar, velocity a vector, as integral of x(t) over time t

Top of page

`"speed"="scalar"=v=d/t`

`"velocity"="vector"= vec v=(Delta vecx)/t rarr `


`vec v=(d vecx)/(dt) rarr `


`int_(x_0)^(x_f)=Delta vec x=dvecx=int_(t_i)^(t_f)vec v dt=vec vint_(t_i)^(t_f) dt=`



`v darr rarr (Delta vec v)`


`vec a=(d vecv)/(dt)`


`vec v=int_(t_i)^(t_f) adt`


`a*t+v_0=v`

`v=v_0+a*t`

`Delta x=v_o*t+a*t^2/2`


`x_(f)-x_0=v_o*t+a*t^2/2`



`x(t)=x_0+v_(o,x)*t rarr t=?`

`t=(x-x_0)/v_(0,x)rarr m/(m/s)=s`


`y(t)=y_0+v_(o,y)*t+g*t^2/2`


`y(x)=y_0+v_(o,y)*((x-x_0)/v_(0,x))+g*((x-x_0)/v_(0,x))^2/2`



`y(x)=y_0+v_(o,y)*((x-x_0)/v_(0,x))+g*((x-x_0)/v_(0,x))^2/2`


`v_(o,y)/v_(0,x)=tan(theta)`


`v_i=sqrt(v_(o,x)^2+v_(o,y)^2`


`y(x=100m)=???`

Calculat the magnitude of the displacement of a car

Top of page

What is the magnitude of the displacement of a car that travels half a lap along a circle that has a radius of `r=150 m`?

`"distance traveled"=|Delta vec x|=|vec x_f-vecx_i|=(2pi*r)/2=pi*r=pi*150m=471.238m `


`"|displacement vector|"=2*r=2*150m=300 m`

How about when the car travels a full lap?

`"distance"=2pi*r=2*471.942.46 m`

`"mag of displacement for a full lap?"`



Solution:
Displacement is the change in position.
Therefore the magnitude of the displacement for half a lap is 300 m .
For a full lap (the car returns to its starting position), the displacement is zero .

`vec v=(vec x_f-vecx_i)/(Deltat)`


`Deltat=t_(f)-t_i=t rarr 0`

`vec v=(dvecx)/(dt) rarr `

`Deltavecx=(vec x_(f)-vec x_i)=intd vecx=int_(t_i)^(t_f)vec vdt`


`x_(f)-x_i=v*t`

`x_(f)=v*t+x_i`

`y=m*x*b`


`v(t=2s)=(8.8m-4m)/(5s-2s)=4.8/3m/s`




`x(t)=5*t^2`


`v_"inst"=(dx)/(dt)=d/(dt)(5t^2)=10*t=f(t)`


`v_"inst"(t=2s)=10*2=20 m/s`

`x_(f)=x_i+v*t`


`v*t=(x_(f)-x_i)*t/t`


`v*t+x_i=x_(f) "eq of motion for motion with constant speed"`



`vec a="acceleration"=(Delta vec v)/(Deltat)`

`vec a=d/(dt)(dx)/(dt)=(d^2x)/(dt^2)`


`"eq of motion for motion with constant acceleration:"`


`x_(f)=x_i+v_i*t+(a/2*t^2) rarr x(t)`


`x(t)=0+0+5*t^2 rarr`


`x_i=0`

`v_i=0`

`a/2=5 rarr a=5*2 =10m/s^2`


`a=(d^2x)/(dt^2)=d/(dt^2)([5*t^2)]=d/(dt)[(10*t)]=10 m/s^2`


`g=9.81 m/s^2 rarr "with each and every second the speed increases by 9.81" m/s`


`"after travelling for 10 sec "rarr "the speed of an object in free fall becomes "98.1 m/s`


`a rarr (m/s)/s=m/s^2`


`a*t+v_i=v_(f)`

`Delta x=x_(f)-x_i=v_o*t+a/2*t^2`

`x_(f)=x_i+v_o*t+a/2*t^2`

Graphical interpretation of speed

Top of page

`a=(Deltav)/t=(v_f-v_i)/(t_(f)-t_i)="slope"="rise"/"run"=(y_2-y_1)/(x_2-x_1)`


`v rarr " along the y/vertical axis"`

`t rarr "along the x/hrizontal axis" rarr`

`"straight line "rarr " the steepness of that line "="acceleration"`

``

`v_f)=a*t+v_i`


`y=m*x+b`

`v rarr" along the y/vertical axis"`

`t rarr "along the x/hrizontal axis" rarr`

`"straight line "rarr " the steepness of that line "="acceleration"`

`v_i rarr " from the intercept, b"`

`b rarr " value of y when x=0"`

Police car catching up with a speeder

Top of page

`v*t=d`


`d=343 m/s*0.5/2s=85.75 m`

`a rarr m/s^2 larr v/tlarr (m/s)/s=m/s*1/s`


`a*t+v_0=v`

`v=a*t+v_0`

`y=m*x+b`

`"velocity"=v rarr "plotted along the vertical/y axis"`

`"time"=t rarr "plotted along the horizontal/x axis"`


`"the line will intercept the y axis at "v_0=b`

`m="slope"=(y_2-y_1)/(x_2-x_1)=(v_(f)-v_0)/(t_(f)-t_o)="rise"/"run"="opposite"/"adjacent"=tan(theta)=a rarr m/s^2`


`"slope"=(Deltav)/t rarr (m/s)/s=m/s^2 "identify which physical property is represented by the slope"rarr m/s^2`

Runners/trains moving towards eact other, calculate after how long they meet

Top of page

`t=d/v rarr s`


`m/(m/s)=m*s/m`

`Delta x=100m=x_L+x_R`

`x_L=?=v_L*t larr v_L=x_L/t`


`x_R=?=v_R*t larr v_R=x_R/t`


`v_L*t+v_r*t=100m`

`t*(v_L+v_R)=100m`


`t=(100m)/((v_L+v_R))`

`t=(100m)/(3.5m/s+4.5m/s)=12.5 s`

`m/(m/s)=m*s/m=s`


`"Verification:"`


`12.5s*3.5m/s+12.5s*4.5m/s=100 m`

`t*v_L+t*v_R=100m`

`x_L=v_L*t=3.5m/s*12.5s=43.75 m`

`x_R=v_R*t=4.5m/s*12.5s=56.25 m`


`x_L+x_R=100m`

Keyword: ;
Submitted 09-01-2021, 13:01  By: coman

`Delta vec x=vec(x_(f)-x_o)`


`"speed"=v="scalar"=|vec v|=(x_(f)-x_o)/(t_(f)-t_i)=(x_(f)-x_o)/(Deltat) rarr m/s`

`Delta t rarr "scalar"`


`vec v=vec(x_(f)-x_o)/t rarr 50 m/s "and" a "direction"`


`y=ax^2`

`m=a*m^2`

`m=1/m*m^2=mrarr 1/m=m^-1`

Converting `g/(cm^3)` into `"kg"/m^3:`

Top of page


Converting `g/(cm^3)` into `"kg"/m^3:`

`200 g/(cm^3)= ?"kg"/m^3`

`200 (g*((1kg)/(1000g)))/(cm*(1m)/(100cm)*cm*(1m)/(100cm)*cm*(1m)/(100cm)) rarr`

`200*(1/100)/(1/(100*100*100))=200*1/1000*(100*100*100)/1= 200*1000"kg"/m^3`
Free fall g=constant, calculate time it takes to hit ground
`g=9.81 m/s^2 rarr"velocity increases by 9.8 "m/s "each and every sec"`


`y(t)=y_0+v_0*t+g*t^2/2`


`y_(f)-y_0=+9.8/2*t^2`

`4m=9.8/2*t^2 rarr`

`t^2=(2*4m)/((9.8m/s^2)/2)=1.632^2 (sec)^2`

`t=sqrt((2*4m)/((9.8m/s^2)/2))=1.632 sec`


`t rarr sqrt(m/(m/s^2))=sqrt(m*s^2/m)=sqrt(s^2)=s`


`t_(1) rarr +, t_2 rarr <0 `

`y(t)-y_0=v_0*t+g*t^2/2`

`4m=2m/s*t+9.8/2m/s^2*t^2`


`4=2*t+9.8/2*t^2`

`t_(1,2)(-b+-sqrt(b^2-4ac))/(2a)`

`v_0=2 m/s`

`t_"thrown object from 4 m"=?`

Distance displacement difference, moving along half circle

Top of page

`"distance"=(2pi*r)/2=pi*r=3.1415*150m=471 m`


`|vec Delta x|=2*r=2*150m=300m`


`"full circle"rarr "distance"=2pi*r=942 m`

`"speed"="magnitude"="scalar"=70 "mi"/"hr"`


`"velocity"="vector"=vec v=70 "mi"/"hr" " northbound"`

Resolving vectors into components

Top of page

`|vec A|=27.9 m`

`sin(theta)="opposite"/"hypothenuse"=A_y/|vec A| rarr`



`A_y=|vec A|*sin(theta)=27.9m*sin(21^0)=9.998 m`

`cos(theta)="adjacent"/"hypothenuse"=A_x/|vec A|`

`A_x=|vec A|*cos(theta)=27.9m*cos(21^0)= 26.04m`



`A^2=A_x^2+A_y^2`

`A=sqrt(A^2)=sqrt(A_x^2+A_y^2)=sqrt(26.04^2+9.998^2)=27.9 m`

`tan(theta)=sin(theta)/cos(theta)="opposite"/"hypothenuse"*"hypothenuse"/"adjacent"="opposite"/"adjacent"=A_y/A_x`

`tan(theta)=A_y/A_x`

`theta=tan^-1(A_y/A_x)=arctan(A_y/A_x)`

`theta=tan^-1(9.998/26.04)=21.0041^0`


`180/pi*tan^-1(9.998/26.04)=21.0041^0`


`180^o/"rad"*"rad"rarr "degrees"`


`sin(theta)=A_y/A`

`cos(theta)=A_x/A`

`A=sqrt(A_x^2+A_y^2`

`tan(theta)=A_y/A_x rarr`

`theta=arctan(A_y/A_x)`

Calculate Components of vectors

Top of page

`|vec a|=15`

`theta_(vec a)=0^o`

`|vec b|=20`

`theta_(vec b)=90^o rarr vecb " has only a vertical component"`

`rarr b_y=sin(90^0)*|vecb|=sin(90)*20=1*20=20`

`a_x=cos(0)*|vec a|=1*15=15`

`|vec c|=|vec a+vec b|=sqrt(a^2+b^2)=sqrt(15^2+20^2)=25`

`theta_("vector sum")=arctan(c_y/c_x)=arctan(b/x)=tan^-1(20/15)=53.13^0`




Keyword: ;
Submitted 09-02-2021, 11:28  By: coman

`vec d`

`d_x=+8.5`

`d_y=-5.5`

`|d|=sqrt(d_x^2+d_y^2)=sqrt(8.5^2+(-5.5)^2)=10.2 m`

`theta=arctan(d_y/d_x)=arctan(-5.5/(8.5))=?`




`theta=55^o`

`|vec F|="magnitude of a vector force"=42 N`

`F_x=F*cos theta=42N*(cos55)=? larr cos theta =F_x/|vecF|`

`F_y=F*sin theta=42N*(sin55)=?larr sin theta =F_y/|vecF|`


Keyword: ;
Submitted 09-02-2021, 15:55  By: coman

`y_(f)=y_0+v_0*t+a*t^2/2 rarr t=?`

`"If thrown with "v_0=2 m/s`



`"+ direction downwards" rarr g=9.8 m/s^2`




`" + direction of our coordinate system downwards"`


`"The origin coincides with initial position"`

`"Choose positive direction in the direction of motion"rarr grarr +`


`y_0=0m`

`g+9.8m/s^2`

`y_f=+10 m`

`v_(0,"downwards")=+2 m/s`

`y_(f)=y_0+v_0*t+a*t^2/2 rarr t=?`

`10=0+2*t+9.8t^2/2 rarr t=?`

`t_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)`


`4.9t^2+2*t-10=0`

`t_2=5/4 s larr " solution"`

`t_1=-5/3 rarr "in the past"rarr "not a physical solution"`




Keyword: ;
Submitted 09-02-2021, 16:57  By: coman

`" + direction of our coordinate system downwards"`


`"The origin coincides with initial position"`

`"Choose positive direction in the direction of motion"rarr grarr +`


`y_0=0m`

`g+9.8m/s^2`

`y_f=+10 m`

`v_(0,"downwards")=-2 m/s`

`y_(f)=y_0+v_0*t+a*t^2/2 rarr t=?`

`10=-2*t+9.8t^2/2 rarr t=?`

`t_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)`


`4.9t^2-2*t-10=0`

`t_2= +5/3=1.666?s larr " solution"`

`t_1=-5/4=-1.25 rarr "in the past"rarr "not a physical solution"`


`t=sqrt(2*h/g)= sqrt(m/(m/s^2))=sqrt(m*s^2/m)=sqrt(s^2)=s`





Keyword: ;
Submitted 09-02-2021, 19:15  By: coman

`l_"classroom"=10.52 m+- 0.005m`

`"age"=?=20 "years" 2" months 3 days 54 minutes "+-60 sec`


`"least count"=1 cm rarr" uncertainty associated with result of measurement"rarr 0.5 cm`

`"least count"=1 mm rarr" uncertainty associated with result of measurement"rarr 0.5 mm =0.05 cm`


`"length"=6.39 m rarr "the 9 is doubtful"`

`"length"=6.39 m+-0.005m`


`6.39 = 6.394=6.389`

`"than 9 is a hundreth of meter"`


`g_"accepted"=9.81 m/s^2 rarr v uarr "by 9.81" m/s "each and every second"`


`g_"measured" ="experimental"rarr "as the slope of a line "=9.72 m/s^2`


`%"error"=|E-A|/A*100=|9.72-9.81|/9.81*100 rarr=0.9174 %`


`15.08=15.1`


`82.453=8.25*10^1`


`11.2cm × 3.4cm × 4.10cm.=?=156.1 cm^3 =?cm^3`


`g=9.81m/s^2`

`"eq of motion for motion with uniform acceleration:"`


`y(t)=y_0+v_0*t+at^2/2`

`v_0=0`

`y_0=0`

`y(t)=g/2*t^2rarr` `y(t) rarr "parabola"`

`g larr"from the slope"rarr`

`"plot y along the verttical axis and "t^2"along the horizontal axis" rarr`

`m="slope"rarr g/2=9.81/2 m/s^2`


`y=m*x+b`

`" How do we linearize a graph"?`

`"m=slope"=g/2 rarr g=2*m=? m/s^2`


`Fuarr=k*xuarr`

`"Given" F(x) rarr k=? ="slope"`

``



Keyword: ;
Submitted 09-03-2021, 12:15  By: coman

`"1 dim motion" rarr " with constant acceleration"`

`"Free fall"rarr a=g=9.81 m/s^2`



`"position as a function of time" rarr "one vertical axis"=y`

`y(t)=y_0+v_0*t+a*t^2/2`

`"drop the rock"rarr v_o=0 m/s rarr v_0*t=0`


`y_0 =0 larr "coveniently choosen"`

`a=g=9.8 m/s^2 rarr "speed increases by "9.8 m/s "each and every second"`

`y=10m=0m+0*t+9.8/2*t^2`


`10m=9.8/2m/s^2*t^2`


`10=9.8/2*t^2 rarr t`

`t^2=2*10/9.8`


`t=sqrt(2*10/9.8)=1.428s="time it takes to hit the ground if rock is thrown"`

`sqrt(m/(m/s^2))=sqrt(m*s^2/m)=sqrt(s^2)=s`


`t=sqrt(2*"height"/"acceleration")`


`y(t)=y_0+v_0*t+a*t^2/2 rarr "choosing the origin at ground level"`


`"choose the positive direction, + direction, upwards"`
`y=10m=10m-9.8t^2/2`

`-10m=-9.8t^2/2 |(*-1)`

`10m=9.8*t^2/2`

`t=sqrt(2*10/9.8)`



`y_f=y(t_"instant it touches ground")=0`


`y(t)=y_(f)=0`

`y_(f)=y_0+v_ot+at^2/2`

`0=10-9.8t^2/2rarr`

`10=9.8t^2/2 rarr t=?`

`t=sqrt(2*10/9.8)`

`y_0=10 m`


`y_(f)=y_0+v_0*t+a*t^2/2 rarr "choosing the origin halfay in between"`

`"+ direction upwards" rarr g=-9.8 m/s^2`

`-5=+5-9.8*t^2/2`

`10=9.8*t^2/2`


`t=sqrt(2*10/9.8)`


`"We can arbitrarly choose the origin and the + direction of our coordinate system"`


`"Simplify your life by choosing the origin to coincide with initial position"`

`"Choose positive direction in the direction of motion"rarr grarr +`


`y=gt^2/2`

`g/2=9.8/2=3.9`

`"in desmos:" y=a*x^2=9.8/2*x^2`

`y=g/2*t^2`

`y=m*x`

`x rarr t^2`

`y rarr y`


`m=g/2 rarr g=2*m=2*"slope"=2*4=8 m/s^2`


`d=2.4 m rarr "leastcount" =1/10"m"`

`"uncertainty"1/2*1/10=0.05m`

`2.4m+-0.05 m`

`2.42=2.44=2.39`

`T=f(L,g)`

`T=f(m,k)`

`s^1=kg^x*((kg)/s^2)^y`



`s^1*kg^0=kg^(x+y)*s^(-2y)`

`0=x+y`

`1=-2y`

`y=-1/2`

`x=?, y=?`

`20cm*(1m)/(100cm)*cm*(1m)/(100cm)*cm(1m)/(100cm)*=20/(100*100*100)=20/10^6=20*10^-6=20E-6=? m^3`


Keyword: ;
Submitted 09-24-2021, 09:32  By: coman

`F_"friction"=mu_k*N`

`N=m*g`


`-m/m*u_kmg(x_2-x_1)=-m/mv_i^2/2`


`u_kg(x_2-x_1)=v_i^2/2`


`(x_2-x_1)=(v_i^2)/(2*u_k*g)`

`(x_2-x_1)=(20m/s)^2/(2*0.8*9.8m/s^2)=25.51mrarr (m^2/s^2)/(m/s^2)=(m^2/s^2)*s^2/m=m`



`W=mv_f^2/2-mv_i^2/2 rarr W="known"=F*Deltax rarr v_i=?`

``

``



`U=mgDeltay rarr J`



`U_"elastic"=kx^2/2=W`


`U+K="constant"`


`DeltaU+DeltaK=0 larr "non conservative forces are are absent"`


`DeltaU+DeltaK+W_"friction"=0`


`Fdarr*tuarr ="constant"`

``

`vec F_"external"=(Delta vecp)/(Deltat) =0 rarr`

`(Delta vecp)=0`


`p_(f)-p_i=0`


`p_f)=p_i`


A cannon `m_"cannon"=1000 kg` fires a shell at a speed `v_"shell"=40 m/s`;

Knowing the mass of the shell `m_"shell"=20 kg` what is the recoil speed of the cannon?


`F_"external"=0`


`p_(f)=p_i`

`p_i=m_"shell"*v_"shell,i"+m_"cannon"*v_"i,cannon"`


`p_(f)=m_"cannon"*v_"f,cannon"`

`m_"shell"*v_"shell,i"=m_"cannon"*v_"f,cannon"`


`m_"shell"/m_"cannon"*v_"shell,i"=v_"f,cannon"`


`v_"f,cannon"=(20kg)/(1000kg)*40m/s=8/10 m/s`



Keyword: ;
Submitted 10-05-2021, 11:47  By: coman

`vec F=m*vec a=(d vecp)/(dt)=d/(dt)(m*vecv)=m*(dvecv)/dt+(dm)/(dt)*vec v=0`

`(dvecp)/(dt)=o rarr dvecp=0 rarr`


`vecp_i=vecp_f`


Big fish, `m_b=4 kg` swallows a small fish initially at rest whose mass `m_s=2 kg`

`"Knowing " v_b=2 m/s, "intial speed of small fish is " v_s=0m/s`

`v_(f,"big+small"=?`


`F_"external"=0 rarr`

`p_i=p_f`

`p_i=m_b*v_b`



`p_f=(m_b+m_s)*v_f`

`(m_b*v_b)/(m_b+m_s)=v_f`

`v_f=(4kg*2m/s)/(4kg+2kg)=1.33 m/s`

`vec p=m*vec v`


`vec F=(dvecp)/(dt)`

`intvecdarr F*dtuarr=vec H="impulse"=intdvecp=Delta vecp`


`v' larr " after collision"`


`v_1 " before collision"`

`p_(i,x,1)`


`p_(f,x,1)`

``

`p_(i,x,2)`



`m_1=1kg`


`m_2=2kg`

`x_1=0`

`y_1=3m`

`x_2=6 m`

`y_2=5 m`


`x_(CM)=(m_1*x_1+m_2*x_2)/(m_1+m_2)=(1kg*0m+2kg*6m)/(1kg+2kg)=12/3=4 m`


`y_(CM)=(m_1*y_1+m_2*y_2)/(m_1+m_2)=(1*3+2*5)/(1+2)=13/3=4.33 m`


``

`X_(CM)=int(vecrdm)/M_"total"`


`rho=m/V rarr dm=rho*dV`

`int(vecrdm)=rhointvecr dV`


`intvec rdxdy`

``

Keyword: ;
Submitted 10-21-2021, 13:15  By: coman

`"Objective"`

Investigate static equilibrium

Determine both experimentally and theoretically the equilibrant force

The force that would balance out the effect of 2 or more forces

Given 2 forces we are going to calculate theoretically the net resultant force of those 2 forces


`F_(x,"resultant/net")=?`

`F_(y,"resultant/net")=?`

`F_("resultant/net")=?=sqrt(F_(x,"resultant/net")^2+F_(y,"resultant/net")^2)`

`theta_"net resultant makes with +x"=arctan(F_(y,"resultant/net")/F_(x,"resultant/net"))=tan^-1(F_(y,"resultant/net")/F_(x,"resultant/net"))`







`"Case A" rarr`

`F_1=5 N " @ 52 degrees with respect to +x, measured counterclokwise "`

`F_1=4 N " @ 120 degrees with respect to +x, measured counterclokwise "`


`"Case B" rarr`

`F_1=2 N " @ 120 degrees with respect to +x, measured counterclockwise " " in the 2nd quadrant" `

`F_1=8 N " @ 310 degrees with respect to +x, measured counterclockwise " rarr " 4th quadrant"`

`"Mention the quadrant that the net resultant force is in!"`

``


`"Sample theoretical calculations; DO not use these numbers but the ones in Case A and Case B":`


`F_1=20 N " @ " 20^o " with respect to +x" `


`F_2=50 N=m*g=m*9.8m/s^2=5 kg*9.8m/s^2 rarr " @ " 50^o " with respect to +x "`


`"Theoretically calculate the x component of the net resultant force"`

`F_(x,"resultant/net")=F_(1,x)+F_(2,x)`

`F_(x,"resultant/net")=20N*cos(20)+50N*cos(50)=50.933 N`

`"Theoretically calculate the x component of the net resultant force"`

`F_(y,"resultant/net")=+-F_(1,y)+-F_(2,y)`

`F_(y,"resultant/net")=20N*sin(20)+50N*sin(50)=45.142 N`

`F_("resultant/net")=?=sqrt(F_(x,"resultant/net")^2+F_(y,"resultant/net")^2)`

`F_("resultant/net")=sqrt(50.9332^2+45.1426^2)= 68.0591N`

`theta=?=tan^-1(45.1426/50.9332)=41.55^o`



Keyword: ;
Submitted 10-28-2021, 09:37  By: coman

`vec g=9.8 m/s^2 larr " speed increases by 9.8"m/s " each and every second" `


`d=r rarr 2d rarr Fdarr " by a factor of " 2^2=4`

`"Calculate the gravitational force exerted by a 1 kg mass upon another 1 kg mass"`

`"1 meter apart"`

`F_g=6.67*10^-11*(1kg*1kg)/(1m)^2=0.0000000000667 N=6.67 *10^-11 N`


``
`g=G*M_"earth"/R_"earth"^2=6.67*10^(-11)*5.9721*10^24/(6371*10^3)^2=9.8138 m/s^2`

`W=100 kg*9.8m/s^2=980 N`

`F_g~1/r^2`


`U~1/r rarr 1/oorarr Urarr 0`


`W=intF*dx=U=int_0^x(G*(m_1*m_E)*x^-2)dx`

`U=G*(m_1*m_E)int_0^x*x^-2dx=G*(m_1*m_E)x^(-2+1)/(-2+1)=U=-G*(m_1*m_E)*1/x`

`v_"esc"=sqrt(2*G*M/R)=11182.46758 m approx 11 "km"/"sec"`


``

If all of mathematics disappeared, physics would be set back by exactly one week. Richard P. Feynman

`F=G*()`

``

---

Keyword: ;
Submitted 09-06-2023, 11:22  By: coman


`vec v=``(d)/(dt) ``(3hat it-4t^2 hat j -2hatk)`


`|vecv|=(3hat i-4*2t hatj)`

`v_(t=2s)=3hati-8*2hatj`


`v_x=3`

`v_y=16 m/s`

`v=sqrt(v_x^2+v_y^2)=sqrt(3^2+16^2
)`

``


Keyword: ;
Submitted 09-06-2023, 11:40  By: coman


`vec v=``(d)/(dt) ``(3hat it-4t^2 hat j -2hatk)`


`|vecv|=(3hat i-4*2t hatj)`

`v_(t=2s)=3hati-8*2hatj`


`v_x=3`

`v_y=16 m/s`

`v=sqrt(v_x^2+v_y^2)=sqrt(3^2+16^2
)`

`vec r=6t^3hatk`


`"which physical property is 6?"`

`s^3*m/s^3 rarr?=? m`

`v(t)=d/(dt)(2thati-10t^2hatj)`


`vec r=2thati-10t^2hatj`

`vec r=v_*t+at^2/2rarr`


`-10=a/2 rarr a=-20 m/s^2`


`tehta=tan^-1(v_y/v_x)=tan^-1(-20/2)=?? "clockwise & counterclocwise +x"`

``

``

---

Keyword: ;
Submitted 09-18-2024, 14:02  By: coman

when elevator moves downwards with `a` and positive direction is downwards

`m_"elevator+people"*g-T=m_"elevator+people"*a`


`m_"elevator+people"*(+9.8)-T=m_"elevator+people"*a`

when elevator moves upwards with `a` and positive direction is downwards

`m_"elevator+people"*g-T=m_"elevator+people"*(-a)`


`m_"elevator+people"*(+9.8)-m_"elevator+people"*(-a)=T`



`T=mgsin(30)=m*a_x`


`F_f=mu*N=??`

`N=???`

Task: write Newton's 2nd law for the block , `m=5kg` pulled up the incline with a force `T=25N` if friction is present, `mu=0.5`.

Calculate acceleration if `theta=30^o`
Choosing`+` in the direction of motion, up he incline,

`T-mgsin(30)-F_f=m_"block"*a_x`


`F_f=mu*N=??`

`N=???`



`"slope"="rise"/"run"=(y_2-y_1)/(x_2-x_1) rarr larr v/t=a`


When choosing a cable that supports an elevator assuming the max no of persons/max weight is `W_"max"=5000 N`.

WHat should be the max tension supported by the cable if max acceleration is `8 m/s^2`.


`T-W=m*a`


`T-m*g=m*a`


`T=mg+ma`

2 masses sitting on 2 inclines


Write Newton's 2nd law along `x` and along `y`


Down the 1st incline is `+`, Newton's 2nd law along x

`W_(1,x)+F_(f,1)-T=m_1*a`


`m_1*g*sin(theta)`


Down the 1st incline is `+`, Newton's 2nd law along x

`W_(1,x)+F_(f,1)-T=m_1*a`


`m_1*g*sin(theta)+mu_1*m*g*cos(theta)-T=m_1*a`

Upward is `+`, Newton's 2nd law along `y`

`N-W_(1,y)=m_1*0`

2nd mass:

`-m_2*g*sin(theta)-mu_2*m_2*g*cos(theta)+T=m_2*a`


``

`T^2/("distance"^3)="constant" rarr ("years")/"km"^3~~1.`

`(T^2/d^3)_("Mars")=1.88^2/1.524^3=0.9987`

1. Weight of person in an elevator moving with acceleration

Top of page

when elevator moves downwards with `a` and positive direction is downwards

`m_"elevator+people"*g-T=m_"elevator+people"*a`


`m_"elevator+people"*(+9.8)-T=m_"elevator+people"*a`

when elevator moves upwards with `a` and positive direction is downwards

`m_"elevator+people"*g-T=m_"elevator+people"*(-a)`


`m_"elevator+people"*(+9.8)-m_"elevator+people"*(-a)=T`

Mass pulled up an incline, tension in cable

Top of page

`T=mgsin(30)=m*a_x`


`F_f=mu*N=??`

`N=???`

Task: write Newton's 2nd law for the block , `m=5kg` pulled up the incline with a force `T=25N` if friction is present, `mu=0.5`.

Calculate acceleration if `theta=30^o`
Choosing`+` in the direction of motion, up he incline,

`T-mgsin(30)-F_f=m_"block"*a_x`


`F_f=mu*N=??`

`N=???`



`"slope"="rise"/"run"=(y_2-y_1)/(x_2-x_1) rarr larr v/t=a`

Mass in an elevator, tension in cable

Top of page

When choosing a cable that supports an elevator assuming the max no of persons/max weight is `W_"max"=5000 N`.

WHat should be the max tension supported by the cable if max acceleration is `8 m/s^2`.


`T-W=m*a`


`T-m*g=m*a`


`T=mg+ma`

2 masses sitting on 2 inclines

Top of page

2 masses sitting on 2 inclines


Write Newton's 2nd law along `x` and along `y`


Down the 1st incline is `+`, Newton's 2nd law along x

`W_(1,x)+F_(f,1)-T=m_1*a`


`m_1*g*sin(theta)`


Down the 1st incline is `+`, Newton's 2nd law along x

`W_(1,x)+F_(f,1)-T=m_1*a`


`m_1*g*sin(theta)+mu_1*m*g*cos(theta)-T=m_1*a`

Upward is `+`, Newton's 2nd law along `y`

`N-W_(1,y)=m_1*0`

2nd mass:

`-m_2*g*sin(theta)-mu_2*m_2*g*cos(theta)+T=m_2*a`


``

`T^2/("distance"^3)="constant" rarr ("years")/"km"^3~~1.`

`(T^2/d^3)_("Mars")=1.88^2/1.524^3=0.9987`

2. Weight of person in an elevator moving with acceleration

Top of page

when elevator moves downwards with `a` and positive direction is downwards

`m_"elevator+people"*g-T=m_"elevator+people"*a`


`m_"elevator+people"*(+9.8)-T=m_"elevator+people"*a`

when elevator moves upwards with `a` and positive direction is downwards

`m_"elevator+people"*g-T=m_"elevator+people"*(-a)`


`m_"elevator+people"*(+9.8)-m_"elevator+people"*(-a)=T`



`T=mgsin(30)=m*a_x`


`F_f=mu*N=??`

`N=???`

Task: write Newton's 2nd law for the block , `m=5kg` pulled up the incline with a force `T=25N` if friction is present, `mu=0.5`.

Calculate acceleration if `theta=30^o`
Choosing`+` in the direction of motion, up he incline,

`T-mgsin(30)-F_f=m_"block"*a_x`


`F_f=mu*N=??`

`N=???`



`"slope"="rise"/"run"=(y_2-y_1)/(x_2-x_1) rarr larr v/t=a`


When choosing a cable that supports an elevator assuming the max no of persons/max weight is `W_"max"=5000 N`.

WHat should be the max tension supported by the cable if max acceleration is `8 m/s^2`.


`T-W=m*a`


`T-m*g=m*a`


`T=mg+ma`

2 masses sitting on 2 inclines


Write Newton's 2nd law along `x` and along `y`


Down the 1st incline is `+`, Newton's 2nd law along x

`W_(1,x)+F_(f,1)-T=m_1*a`


`m_1*g*sin(theta)`


Down the 1st incline is `+`, Newton's 2nd law along x

`W_(1,x)+F_(f,1)-T=m_1*a`


`m_1*g*sin(theta)+mu_1*m*g*cos(theta)-T=m_1*a`

Upward is `+`, Newton's 2nd law along `y`

`N-W_(1,y)=m_1*0`

2nd mass:

`-m_2*g*sin(theta)-mu_2*m_2*g*cos(theta)+T=m_2*a`


``

`T^2/("distance"^3)="constant" rarr ("years")/"km"^3~~1.`

`(T^2/d^3)_("Mars")=1.88^2/1.524^3=0.9987`

elevator moves downwards with `a`

Top of page

when elevator moves downwards with `a` and positive direction is downwards

`m_"elevator+people"*g-T=m_"elevator+people"*a`


`m_"elevator+people"*(+9.8)-T=m_"elevator+people"*a`

when elevator moves upwards with `a` and positive direction is downwards

`m_"elevator+people"*g-T=m_"elevator+people"*(-a)`


`m_"elevator+people"*(+9.8)-m_"elevator+people"*(-a)=T`

Newton's 2nd law for the block , `m=5kg` pulled up the incline

Top of page


`T=mgsin(30)=m*a_x`


`F_f=mu*N=??`

`N=???`

Task: write Newton's 2nd law for the block , `m=5kg` pulled up the incline with a force `T=25N` if friction is present, `mu=0.5`.

Calculate acceleration if `theta=30^o`
Choosing`+` in the direction of motion, up he incline,

`T-mgsin(30)-F_f=m_"block"*a_x`


`F_f=mu*N=??`

`N=???`



`"slope"="rise"/"run"=(y_2-y_1)/(x_2-x_1) rarr larr v/t=a`


When choosing a cable that supports an elevator assuming the max no of persons/max weight is `W_"max"=5000 N`.

WHat should be the max tension supported by the cable if max acceleration is `8 m/s^2`.


`T-W=m*a`


`T-m*g=m*a`


`T=mg+ma`

2 masses sitting on 2 inclines


Write Newton's 2nd law along `x` and along `y`


Down the 1st incline is `+`, Newton's 2nd law along x

`W_(1,x)+F_(f,1)-T=m_1*a`


`m_1*g*sin(theta)`


Down the 1st incline is `+`, Newton's 2nd law along x

`W_(1,x)+F_(f,1)-T=m_1*a`


`m_1*g*sin(theta)+mu_1*m*g*cos(theta)-T=m_1*a`

Upward is `+`, Newton's 2nd law along `y`

`N-W_(1,y)=m_1*0`

2nd mass:

`-m_2*g*sin(theta)-mu_2*m_2*g*cos(theta)+T=m_2*a`


``

`T^2/("distance"^3)="constant" rarr ("years")/"km"^3~~1.`

`(T^2/d^3)_("Mars")=1.88^2/1.524^3=0.9987`




`F_p-W_x-F_f=m*a_x`

`W_"net"=F_"net"*x`


`W_x=mgsin(theta)`

`F_f=mu*N=mu*mgcos(theta)`

---

Keyword: ;
Submitted 09-11-2023, 13:09  By: coman

`x(t)=x_i+v_i*t+at^2/2`



`vec r(t) rarr vecv =(dvecr)/(dt)`


`v_(0,y)=5 m/s`


`v_(0,x)=2 m/s`

`vec a=(dvecv)/(dt)`

`dvec v=int vecadt`


`vec v=``5 hat i`` + ``2 hat j``+int _0^t ``(3thati +4 t hatj)dt`

`vec v=``5 hat i`` + ``2 hat j``+ ``(3t^2/2hati +2 t^2 hatj)`

`vec r=``vec r_(o)+``int _0^(t_f) vec vdt=int5 hat i`` + ``2 hat j``+ ``(3t^2/2hati +2 t^2 hatj)dt`

`r(t)=r_i+v_ot+at^2/2`

``


`rho=m/V`

`a*t+v_i=(v_f)`


`x=160+4*120=640m`


`"Officer moves with acceleration a"`


`a=2m/s^2=`


`x_p(t)=x_0+v_0t+a*t^2/2`


`x_s=8*t`


`x_p(t)=2*t^2/2`

`8t=2t^2/2`

`t=?`

``

Keyword: ;
Submitted 09-13-2023, 10:56  By: coman

`"along y" rarr W=N`

`m*g=N`


`"along x"`


`F_"applied"-F_(f)=m*a`

`F_f=mu*N=mu*m*g`


`a=(F_"applied"-mumg)/m`


`"along y mass on incline, "theta rarr W_y=N`


`W_y=W*cos(theta)`

`"along x mass on incline, "theta`


`W_x-F_(f)=m*a_x`


`(W_x-muN)/m=a_x`



`(W_x-F_(f))/m=a_x`

`(W_x-muW_y)/m=a_x`


`(m*g*sin(theta)-mu*m*g*cos(theta))/m=a_x`


`(g*sin(theta)-mu*g*cos(theta))=a_x`

`g*(sin(theta)-mu*cos(theta))=a_x`


`mu=F_f/N, theta=20^o`


`"taking that to the extreme, to the limit" rarr theta=90 rarr a_x=9.81 m/s^2?`

`mu=0.3 rarr `

`g=9.81;theta=20;mu=0.3;a_x=g*(sin(theta)-mu*cos(theta))=0.5897`

Investigating Ohm's Law, experiment

Top of page

`"Objective: Investigating Ohm's Law"`

`V=I*R`

`R=V/I rarr Omega=V/A`


`"slope"="rise"/"run"`

`%"error"=|M-A|/A*100 rarr %`

`M rarr "slope"`


`A rarr"set your R to"`

`R > 10,000 Omega`

Answer:

1. What was the objective?

2. What were the physical properties directly measured?

3. What were the physical properties determined/calculated based on the measured ones?

4. What were the physical properties directly measured?

5. What does the slope represent? (Which physical property?)

6. What are the units for the slope?

7. Calculate percent error. Does it have units ?

Momentum conservation

Top of page

`vecF=(dvecp)/(dt)rarr " if ext forcees=0"rarr dvecp=0rarrp_f)=p_i`


`vec p=m vecvrarr kg*m/s`


Big fish `M=10 kg` , eats/swallows small fish, `m=2 kg`.

`V=2m/s` small fish is at rest`v=0m/s`

Calculate speed of big fish after eating that big lunch.

`p_(f)=p_i`


`p_i=M*V+m*0`

`p_F=(M+m)*v_f`


`M*V=(M+m)*v_f`


`M*V/(M+m)=v_f`


`10*2/(10+2)=v_f>


`m_b*V_b+(-M_R*v_R)=0`

`vecF*dt=dvecp=H`

`N*s=kg*m/s^2*s=kg*m/s`

Center of mass, 3 point particles

Top of page


`x_"CM"=((-2)*4kg+4*8kg+1*4kg)/(4+8+4)`


`y_"CM"=((3)*4kg+2*8kg+(-2)*4kg)/(4+8+4)`


`a_"net"=F_"net"/(m_1+m_2+m_3)`


`F_"net"=sqrt(F_x^2+F_y^2)`


`a_"net"=sqrt(a_x^2+a_y^2)`


`theta=ta^-1(F_y/F_x)=tan^-1(a_y/a_x)`

Momentum conservation collision of 2 point partticles

Top of page

`p_i=p_f`


`m_1*v_1+m_2*v_2=m_1*v_1'+m_2*v_2'`

Collision `m_1=2000`, `m_2=?`

`v_1=10m/s`

`v_2=-20m/s`

`v_(f)=0`


`2000kg*10m/s-m_2*20=0`

`(2000kg*10m/s)/20=m_2=20000/20=1000 kg`


`p_x=0`

`p_y=0`


`m_1*v_(1,x)+m_2*v_(2,x)-m_3*v_(3,x)=0`

`-m_1*v_(1,y)-m_2*v_(2,y)+m_3*v_(3,y)=0`

momentum conservation along x:


`m_1=10, m_2=20, m_3=40`

`10*(-v_1)*cos(theta_1)+20*(v_2)*cos(theta_2)+10*(v_3)*cos(theta_3)=0`
``


momentum conservation along y:

`10*(-v_1)*sin(theta_1)+20*(v_2)*sin(theta_2)+10*(v_3)*sin(theta_3)=0`

---

Task 1: Momentum Conservation: riffle,

Top of page

Use momentum conservation to explain what would happen in this situation, and why!

Riffle Recoil, momentum conservation

Task 2: Momentum Conservation: jellyfish

Top of page
Use momentum conservation to explain how does a jellyfish propels itself forward. Did a jellyfish take a physics course?
Does the same explanation apply to a rocket?

Task 3: Using the momentum conservation principle calculate the speed of a big fish, `v_"big"` , `m_"big"=10 kg` after swallowing a small fish `m_"small"=2 kg` initially at rest!

Top of page
`p_"initial"=p_"final"`

---

Translational Rotational analogues:

Top of page
`m rarr I="moment of rotational inertia"` `x rarr theta`
` vec v rarr vec omega="angular speed"`
`veca rarr vec alpha =" angular acceleration"`
` vec F rarr vec tau="torque, external influence that tends to change the state of rotational motion"`
`vecp rarr vec L="angular momentum"`

Using the analogy -above-between physical properties used to describe translational motion and physical properties used to describe rotational motion rewrite Galielo's formula,

`v_"final"^2=v_"initia"^2+2*a*Deltax`



Using the analogy between physical properties used to describe translational motion and physical properties used to describe rotational motion write kinetic energy due to rotational motion.

`K_"translational"=1/2*m*v^2`

phy2053-rolling-down-incline

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https://www.geogebra.org/m/ztUKm6QV Moment of Inertia: Rolling and Sliding Down an Incline Author:Tom Walsh Topic: Rotation

`vecL=vec r times vec p=vec r times m vecv`

`vec L=I vec vec omega`

`vec v=r vecomegararr m/s=m*"rad"/"sec"`

`vectau= vec r times vecFrarr N*m`

`vec tau=(dvecL)/(dt) =0 rarr`

`dvecL=0`

`L_i=L_f`

`I=mr^2rarr uarr`

`I_"arms are pulled in" < I_"arms are stretched out"`

`I_"arms are stretched out"*omega_"arms are stretched out"=I_"arms are pulled in"*omega_"arms are pulled in"`

`K=m*v^2/2`

`K_"rotational"=I*omega^2/2`

`mgh=K_"tr"+K_"rot"`

`"sphere"`

`mgh=m_"sphere"*v_"sphere"^2/2+I_"sphere"*omega_"sphere"^2/2`

`mgh=m_"sphere"*v_"sphere"^2/2+2/5mr_"sphere"^2*v^2/ (2*r_"sphere"^2)`

`gh=*v_"sphere"^2/2+2/5r_"sphere"^2*v^2/ (2*r_"sphere"^2)`

`gh=v_"sphere"^2/2+2/5*v_"sphere"^2/ (2)`

`gh=1/2*v_"sphere"^2+1/5*v_"sphere"^2`

`gh=5/10*v_"sphere"^2+2/10*v_"sphere"^2`

`gh=7/10*v_"sphere"^2`

`v_"sphere"=sqrt(10/7gh)`

`I_"sphere"=2/5mr_"sphere"^2`

`v=r*omegararr omega^2=v^2/r^2`

`mgh=m_"shell"*v_"shell"^2/2+2/3mr_"shell"^2*v^2/ (2*r_"shell"^2)`

`mgh=m_"shell"*v_"shell"^2/2+2/3mr_"shell"^2*v^2/ (2*r_"shell"^2)rarr `
`v_"shell"=?<>v_"sphere"??`

`gh=1/2v_"shell"^2+1/3*v^2 rarr `

Task: rotational motion, satellites, does `omega_"earth "` increase or decrease as you send a satellite into orbit?
Hint 1: do you permanently remove mass from the Earth's surface?
Hint 2: as you remove mass from the Earth's surface, does its moment of inertia decrease or increase?
Hint 3: if angular momentum is conserved:
`L_("final")=L_"initial"`

`I_("final")*omega_("final")=L_"initial"*omega_("initial")`

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Task: rotational motion, satellites, does the period of revolution `T` increase or decrease as you send a satellite into orbit? Top of page

Hint 1:
`omega_"earth "=2*pi*"frecquency"=2*pi/"period"=2*pi/T`

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Task: does a sphere have a greater speed at the botom of an incline than a sperical shell, if they roll down the incline without slipping, from the same height `h`?
Calculate their kinetic energies in terms of `g` and height, `h` they are released from..

`m_"sphere"=m_"shell"`

moment of inertia:
`I_"sphere"= 2/5*m_"sphere"*R_"sphere"^2`

`I_"shell"= 2/3*m_"shell"*R_"shell"^2`

Hint 1: total kinetic energy is the sum of translational kinetic and rotational kinetic:
`K_"total"=K_"translational"+K_"rotational"`

`K_"total"=1/2*m*v_"sphere"^2+1/2*I_"sphere"*omega_"sphere"^2

Hint 2:

Linear velocity `v`, is related to angular velocity, `omega` through the radius of circular mortion, `R`:

`v=omega*R`

Rotational static equilibrium, torques Experiment

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`m_"40 cm"=m_"total"*40/100`

`tau_"40 cm"=m_"40 cm"*50/100*9.8`

`tau_"100g"=100/1000*L_1*9.8`

`m_"40 cm"*50/100*9.8=100/1000*L_1*9.8`

`L_(1,"theoretical")=?`

`L_(1,"experimental")=?`

`%"error"=?`

Task torques, simple

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`vec tau=vec r times vec F`

`tau= r* F *sin(theta)`

Which angle is given in that parallelogram?

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Tasks fluids pressure, Pascal's pronciple

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Buoyant force definition:

`F_b=W_"fluid displaced"=m_"fluid displaced"*g=`

`F_b=rho_"fluid"*V_"fluid displaced"*g`

`F_b=W_"fluid displaced"=m_"fluid displaced"*g=`

`rho_"fluid"=m/V rarr`

` m_"fluid displaced"=rho_"fluid"*V_"fluid displaced"`

`F_"buoyant"=rho_"fluid"*V_"fluid displaced"*g="Weight"_"to be lifted up"="mass"_"to be lifted up"*g`

`rho_"fluid"*V_"fluid displaced"="mass"_"to be lifted up"`

`V_"balloon"=V_"fluid displaced"`

`V_"fluid displaced"="mass"_"to be lifted up"/rho_"fluid"`

Task 1: Calculate the volume of a balloon such that it lifts up a `100 kg ` mass.

Ignore the weight of the balloon's skin and the weight of the gas inside the balloon. `rho_"air"=1.225 "kg"/m^3`

Task 2: How would object in an airplane move if the window were to break? From inside out ?

What do alligators and submarines have in common?

`rho_"submarine"=(m uarr)/(V)`

Task 3: how do fish use the definition of density to resurface ? Hint: if they inflate their swimming bladder what happens to their density? How much would a block be submerged in water if

`h_"block"=4 m`

`rho_"block"=700 "kg"/m^3`

`"length"="width"=5 m`

`F_"buoyant"="Weight"`

`W_"fluid displaced"=W_"block"`

`m_"fluid displaced"*g=m_"block"*g`

`rho_"fluid displaced"*V_"fluid displaced"*g=rho_"fluid displaced"*"Area"_"block"*h_"block submerged"*g`

`h_"block submerged"=2.8 m`

`h_"block"=4 m`

Specific density:

`rho_"block"/rho_"water"=(700 "kg"/m^3)/(1000"kg"/m^3)=0.7 rarr 70% "of block is under water"`

`(2.8m)/(4m)= 0.7rarr 70% "of block is under water" `

Task 4: If a rectangular block has a density of `rho_"block"=800 "kg"/m^3` and is placed in a fluid whose density is `1200 "kg"/m^3` what percentage of the block is under fluid?

Hint: use specific density formula:

`rho_"block"/rho_"fluid"rarr *100 rarr %`

Task 5: Calculate radius of piston in a hydraulic lift system if the diameter of the large piston is `100 cm^2` and you want to lift up a `40,000 N` weight with your own weight= `m_"student"*g`.

Use your own mass.

Pascal principle, sample calculation:

lifting up a weight of `20,000 N` with your own weight `100kg*9.8m/s^2~~1000 N`

`F_u/A_u=F_d/A_d`

`F_u/r_u^2=F_d/r_d^2`

`F_u=W_"car"=2000kg*9.8m/s^2~~20,000N`

`F_d=100kg*9.8=1000N`

`r_u^2/r_d^2=F_u/F_d=(20,000N)/(1000N)~~20`

`pdarr*Vuarr=nRT="constant"`

`Delta p=rho*g*y`

Hubble's Law

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`67±4 ("km"/"sec")/"Megaparsec"`
`v_"recessional"=H*d`
H=Hubble's constant

Archimede's principle,buoyant force experiment

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`rho=m/Vrarr "kg"/m^3`
`F_b=m_"water displaced"*g`

`F_b=rho_"fluid"*V_"water displaced"*grarr N`
`F_b=W_"air"-W_"water"`

`rho_"accepted brass"=?`
`rho_"measured brass"=?`
`rho_"accepted steel"=?`
`rho_"measured steel"=?`
`rarr % "error of densities"=?`

`F_"b upon steel"=W_"air"-W_"water"=?`

`F_"b upon brass"=W_"air"-W_"water"=?`

`F_"b upon brass"=rho_"fluid"*V_"water displaced"*grarr=? N`

`F_"b upon steel"=rho_"fluid"*V_"water displaced"*grarr =? N` `` text

`% "error of buoyant force"=?`



Calculate the height of Hg in a mercury/hydrargirium barometer:

Only physics: weight of Hg in the tube is balanced out by force due to atmospherric pressure=101,000 newtons per square meter!

`p=F/Ararr F=p*A`

`F=p*A=101,000 N/m^2*pi*r^2`

`W_"Hg"=m_"Hg"*g=rho_"Hg"*V_"Hg"*g`

`101,000*A_"tube"=rho_"Hg"*V_"Hg"*g`

`101,000*A_"tube"=rho_"Hg"*A_"tube"*h_"Hg"*g`

`(101,000*A_"tube")/(rho_"Hg"*A_"tube"*g)=h_"Hg"`

`(101,000)/(13,600 (kg)/m^3*9.8 m/s^2)=h_"Hg" rarr "meters"`

`h_"Hg"=(101000)/(13600*9.8)= 0.7578 m=75.78 cm=757.8 mm`

Task 2: Calculate the height of water in a `H_2O` barometer

Simple harmonic motion/oscillator:

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Links:
circularSHM.swf
circularSHMvectors.swf
SHMPeriodAmplitude.swf
shm_Oscillations_Algebra-based.htm
shm_Oscillations_Algebra-based.htm

shm_Oscillations_Algebra-based.htm
http://hyperphysics.phy-astr.gsu.edu/hbase/shm.html
Geogebra unit circle

Geogebra unit circle
Geogebra unit circle 2


What we want is equation of motion for a simple harmonic oscillator:
`x(t)=?`

`F_"elastic restoring"=-k*x`

`m*(d^2x)/(dt^2)=-kx`

`(d^2x)/(dt^2)=-k/m*x`

Let's look at the units for `k/m=(F/x)/mrarr =((kg*m/s^2)/m)/(kg)=1/s^2 rarr`

`sqrt(1/s^2)=1/srarr omegararr "rad"/"sec" rarr`

`sqrt(k/m)=omega rarr`

`(sqrt(k/m))^2=omega^2`

`(d^2x)/(dt^2)=-omega^2*x `

Task 3: verify that the below is the solution to the 2nd order diff eq above

`x=A*sin(omega*t)=A*sin(theta)`

`d[(A*sin(omega*t))]/(dt)=?=Aomega*cos(omega*t)`

`d(omega*cos(omega*t))/(dt)= -Aomega^2*sin(omega*t)` `` `` Top of page

Force needed to completely submerge a beach ball, downward force acting against upward buoyant force

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Calculate the force exerted upon a `r=0.5m` radius beach ball to completely submerge it in water!

`F_"buoyant"=F_"exerted`

`F_"buoyant"=rho_"fluid"*V_"fluid displaced"*g`

`V_"fluid displaced"=V_"beach ba;;"=4/3*pi*"radius"^3`

Bernoulli's principle

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Product between area of pipe and velocity of fluid is a constant!
Inverse proportionality rlationship betwee cross sectional area, `A=pi*"radius"^2` and velocity of fluid :
`Auarr * v_"fluids"darr="constant"`
Constant rate of flow, look at units of area times units of velocity:

`m^2*m/s=?`

`A_"capillaries"*v_"capillaries"=A_"aorta"*v_"aorta"`

`A_"capillaries"*v_"capillaries"/v_"aorta"=A_"aorta"`

`A_"aorta"=pi*r_"aorta"^2 rarr "units=?"`

`m^2*m/s=m^3/s="flow rate"`

Simple Harmonic motion: tasks, units

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Task 1: show that angular frequency of an SHM/pendulum is expressesd in `"rad"/"sec"rarr 1/s=Hz`

`omega=sqrt(g/L)`

`g rarr m/s^2`

`L="length of string"rarr m`

Task 2: show that units for the speed of a transverse wave, `v=sqrt("tension"/"linear mass density"` are `m/s`

tension is a force `rarr N=(kg*m/s^2)`

`"linear mass density"="mass"/"length" rarr (kg)/m`

`x(t=0) = (5cm)*cos(pi/6)`



`v(t=0) = omega*(5cm)*cos(pi/6)`

`x(t=0) = (A)*cos(omega*t+phi)`

`v(t=0) =(dx)/(dt)= -omega*(A)*sin(omega*t+phi)`

`v_"max" larr sin_"max"rarr (omegat+phi)=pi/2 `

`a_"max"=A-omega^2*cos(omegat+phi) larr cos_"max"rarr (omegat+phi)=? `

`E_"SHM"=k*A^2`

`W=int_(x_i)^(x_f)F(x)dx=int_(x_i)^(x_f)(-k)xdx=-kx^2/2=U`

`omega=sqrt(k/m)=2pif=2pi/T`

`(sqrt(k/m))^2=(2pif)^2=(2pi/T)^2`

`k/m=4pi^2*f^2=4pi^2/T^2`

`v=lambda*f=lambda*1/T`

`e^(alpha*x)=2.71^(alpha*x)`

Doppler effect

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`f_"observer"=f_"source"*(v_"soundwave")/(v_"soundwave"+-v_"source")`

`f_"observer"
`f_"observer when source is moving away"=f_"source"*(v_"soundwave")/(v_"soundwave"+v_"source")`

A sound source is moving at `12 m/s`, `f=1200 Hz `toward a stationary listener that is standing in still air. (a) Find the wavelength of the sound in the region between the source and the listener. (b) Find the frequency heard by the listener.

Solution: `lambda``=``(v_"sound"+-u_"source")/(f_"source")`

`lambda=(345m/s-12 m/s)/(1200 1/s) rarr (m/s)/(1/s)=v=0.27 m=m/s*s/1`

frequency at the listener:

`f_"receive/observe"``=``(f_s)*(v+-u_r)/(v+-u_s)`

`u_r` speed of the receiver , `u_s` speed of the source

Converting intensity from decibells into `"Watts"/m^2`

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`I_"waves"=P/A=(E/t)/A=E/(t*A)rarr "Joules"/(sec*m^2)rarr 1300`

`I_"threshold"=10^-12 W/m^2 rarr "Watt"=(1"joule")/"sec"`

`"intensity="W/m^2 rarr dB`

`I("in dB")=10*log_10(I_("in"W/m^2)/I_0)`

`I("in dB corresponds to" 10^-10)=10*log_10((10^-10)/10^-12)=10*log_10(10^2)=10*2=20 db`

`I("in dB corresponds to" 10^-9)=10*log_10((10^-9)/10^-12)=10*log_10(10^3)=10*30=30 db`

10*log(10^-5/10^-12)

10*log(10^-6/10^-12)

Calculate the intensity in `W/m^2` of a sound whose intensity in `dB` is `70 dB.`

`I_"in dB"=10*log_10I_("in"W/m^2)/I_"threshold"`

`10^((70 dB)/10)=I_("in"W/m^2)/10^-12"`

`10^7*10^-12=10^-5W/m^2=I_("in"W/m^2)`

`e^x=12`

`x=ln12`

`x_1,x_2,x_3,c*t`

`log_10(10^-10/10^-12)=log_10(10^2)=2log_10 10=2*1`

`a^x=b rarr x=log_ab`

`d_"obstacle"=v_"sound built in"*t-v_"flying bat"*t_"in flight"`

Task: speed of sound as a function of temperature, killing bats

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The speed of sound in a gas is given by `v = sqrt(gamma(RT)/M)` where `R` is the gas constant, `T` is the absolute temperature,

`M` is the molecular mass of the gas, and `gamma` is a constant
that is characteristic of the particular molecular structure of the gas.
If temperature `T` increases `4` fold by what factor does speed increase?
A) 4
B) 3
C) 2
D) 1

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In order do kill all bats in a cave, assuming speed of sound emiitted by bats is buil in into their brains:
1. Do you have to increase or decrease the temperature `T`?
Hint: in order to kill bats they have to crash into the wall.
Do they have to overestimate or underestimate the distance to the wall ?

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