Practice Test 7 Static Gravitation Shm

7/28/2019 Practice Test 7 Static Gravitation Shm

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AP Physics Practice Test:Static Equilibrium, Gravitation, Periodic Motion

2011, Richard White www.crashwhite.com

This test covers static equilibrium, universal gravitation, and simple harmonic motion, with some problemsrequiring a knowledge of basic calculus.

Part I. Multiple Choice

1.

A mass of 10 kg is suspended from a cableA and a rigid horizontal bar B that is free to rotate, as shown.What is the approximate tension, in Newtons, in cableA?

a. 1003

b. 100 3 c. 200

2

d. 200 3 e. 200

3

2. In a science fiction story, a planet has half the radius of the Earth, but the same mass as the earth. What isthe acceleration due to gravity at the surface of this planet as a function ofg?

a. 4gb. 2g

c. gd. 1

2g

e. 14g

60

10kg

A

B


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3.

The pulley system consists of two solid disks of different radii fastened together coaxially, with two differentmasses connected to the pulleys as shown above. Under what condition will this pulley system be in staticequilibrium?

a. m=Mb. rm= RMc. r2m= R2Md. rM= Rme. r2M= R2m

4. Earth and Jupiter both travel in a roughly circular orbit around the sun. Jupiters orbit is approximately 5times the radius of the Earths orbit. What is the approximate relationship between the centripetalacceleration of each planet?

a. aEarth = aJupiterb. aEarth = 5 aJupiterc. aEarth = 25 aJupiterd. aJupiter = 5 aEarthe. aJupiter = 25 aEarth

5. A simple harmonic oscillator consisting of a massMattached to a spring with spring constant k is set intomotion at the surface of the earth, and observed to have a frequencyf. The same spring is then attached to amass of 2M, and moved to a location Rabove the surface of the earth, where Ris the radius of the earth.What is the frequency of oscillation now?

a. f b. 2f c.

4f

d. f2

e. f2

M m

Rr


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6. A particle moves constantly in a circle centered at the origin with a period of 4.0 seconds. If its position attime t= 0 seconds is (2,0) meters, two possible equations describing the particles x-andy-components are:

a. x = 2cos 2

t!

"#

$

%& y = 2sin

2t

!

"#

$

%&

b. x = 2cos 2

t!

"#

$

%& y = 2sin

2

t!

"#

$

%&

c. x = 2sin 2

t!

"#

$

%& y = 2cos

2t

!

"#

$

%&

d. x = 2sin 2

!"# $

%& y = 2cos

2!"# $

%&

e. x= 2 cos 2t( ) y = 2sin 2t( )


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Part II. Free Response

7.

A long, uniform beam with mass mand length Lis attached by means of a pivot, located at L/4, to a verticalsupport as shown above. The beam is connected to a support line oriented at an angle relative to thehorizontal; the tension in the support line is indicated by a Force sensor of negligible mass. The beam iscurrently oriented in a horizontal position. Give all answers in terms of variables given and fundamentalconstants.

a. Show that the moment of inertia for the rod about the pivot is 748

ML2 .

b. Draw a free-body diagram of the horizontal beam.

1

4L

3

4L

springscale

cm m


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A weight of mass 2mis now placed at the left end of the beam, and a mass of 3mis located at the right endof the beam.

c. Calculate the tension Tin the support line.

d. Calculate the xandycomponents of force acting on the pivot point.

cm m

2m3m


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e. The support line for the beam is now cut. Calculate the angular acceleration of the beam at themoment the support line is cut.

8. A satellite with mass m = 617kg is placed into a circular orbit 1.00107m above the surface of the earth,

which has a mass M= 5.981024kg and a radius r= 6.38106m .

a. Calculate the linear speed of the satellite as it orbits the earth at this altitude.

b. Calculate the radial acceleration of the satellite.


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c. Calculate the period of the satellite.

d. Calculate the satellites total mechanical energy at this altitude.

e. Determine the escape velocity that was necessary for the satellite to be launched into this orbit fromthe surface of the earth.

f. Calculate the location of the center-of-mass of the earth-satellite system, relative to the center of theearth.


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9.

A spring of negligible mass and spring constant 500 N/m is oriented vertically as shown. A flat 1.00 kgplatform is then attached to the top of the spring.

a. What distance does the spring compress when the 1.00 kg platform is attached and slowly allowed todescend to an equilibrium position?

The platform is now pulled down an additional 10 centimeters from the equilibrium position and released.b. Calculate the period of the mass-spring systems oscillation.

c. Determine the speed of the platform as it passes the equilibrium position during its oscillation.

k=500N/m

m=1.00kg


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Now the platform is placed again at its equilibrium position. A 500 gram blob of clay is released from aheight of 30.0 centimeters above the platform so that it falls and contacts the platform in a perfectly inelasticcollision.

d. Calculate the period of this new mass-spring system.

e. Calculate the amplitude of this new mass-spring system.

k=500N/m

m=1.00kg

h=30.0cm

m=0.50kg


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The same experiment is now conducted by astronauts at a base station on Mars, where the acceleration dueto gravity is less than it is on earth. The 500-gram blob of clay is again released from a height of 30.0centimeters above the platform, and it falls and collides perfectly inelastically with the platform as before.

f. Compared with the experiment on earth, the period of the mass-spring system on Mars is more, less,or the same?Explain.

g. Compared with the experiment on earth, the amplitude of the mass-spring system on Mars is more,less, or the same?Explain.


11/17

AP Physics Practice Test Solutions:Static Equilibrium, Gravitation, Periodic Motion


1. The correct answer is e . The weight of the mass, approximately 100N, must be entirely supported by thevertical component of the tension in the cable, Fy . Therefore:

Fy = maFy Fg = 0

Fy = FTension sin60 = (10kg)(~ 10m /s2

)

FTension =100N

3 2= 200 3

2. The correct answer is a . The acceleration due to gravity can be determined using Newtons Law ofUniversal Gravitation:

Fg =Gm

1m

2

r2

m1g =G m1m2r2

g =Gmearth

r2

Now, determine agravity for the new planet:

ag =Gmplanet

rplanet2

=

Gmearth1

2rearth( )

2=

4Gmearth

rearth2

ag = 4g

3. The correct answer is d . The pulley system will be in equilibrium when the sum of the torques acting onthe pulleys is 0.

= 0M m = 0

= rF

rMg( ) Rmg( ) = 0

rMg =Rmg

rM= Rm


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4. The correct answer is c . The centripetal acceleration of each planet is driven by the force of gravity, andthe acceleration of a planet can be calculated as follows:

Fg=G

Mm

r2

Fg=

mag

mag =

GMm

r2

ag =GM

r2

We can see that acceleration due to gravity is inversely proportional to the square of the radius. Jupiter, withits radius 5 times that of the Earth, has 1/52, or 1/25, the acceleration of the Earth. This relationship isconsistent with answer c.

5. The correct answer is e. The location of the mass-spring system doesnt have any effect on its frequency

of oscillation, but the mass 2Mattached to the spring does:

fmassspring =1

2

k

m

$f =1

2

k

2m=

f

2

Note that changing the location of a pendulum doesaffect the frequency of its oscillation, according to theequation

fpendulum =1

2

L

g

6. The correct answer is a . Based on the circular path being centered at the origin and particles originalposition at (2,0) meters, we can deduce that the amplitudeA of the particles motion is 2 meters. Its periodof 4.0 seconds allows us to determine its angular velocity:

T =2

=2

T=

2

4=

2

Equations that describe the components of circular motion are:x = Acos(t+); y = Asin(t+)

Here, substituting in the values from above, and with = 0, the possible equations are:

x = 2cos

2t

#

$%

&

'(; y = Asin

2t

#

$%

&

'(

Note that we dont know whether the particle is moving in a counterclockwise or clockwise direction; theyequation is one possible solution (for a ccw motion). A clockwise motion would have the same equationdescribing the xcomponent, but theycomponent would be given by y = Asin(t+) , or

y = 2sin

2t

"

#$

%

&' .


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7.

a. There are two ways to arrive at the moment of inertia for the rod: perform the full I= r2 dmintegration with limits (-1/4)Lto (+3/4)L, or use the parallel-axis theorem, shown here.

I= Icm+ MD

2

I=1

12ML

2+ M

L

4

!

"#

$

%&

2

=

7

48ML

2

b.

c. This is a static equilibrium problem where the sum of the torques acting on the beam is zero. = 0

2m m 3m +Tension = 0

rFg2m rFg rFg3m + rT= 0

L

4

$

%&

'

()2mg

L

4

$

%&

'

()mg

3

4L

$

%&

'

()3mg+

3

4L

$

%&

'

()Tsin= 0

T=8mg

3sin

d. To determine the Rx and Rycomponents,well use Newtons Second Law inhorizontal and vertical directions.Horizontally:

Fx = 0Rx Tx = 0

Rx =Tx =Tcos=8mg

3tan

Vertically:Fy = 0

Ry+T

y F2mg Fg F3mg = 0

Ry = 2mg+mg+3mgTsin

Ry = 6mg8mg

3sin

#

$%

&

'(sin=

10

3mg

T

FgRx

Ry

T

Fg

Rx

Ry


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e. To determine angular acceleration, use the angular form of Newtons Second Law of Motion. Notethat the moment of inertia of the system has increased with the two masses.

I= Ibeam

+I2m+ I

3m=

I= Ibeam

+M2R

2

2+ M

3R

3

2

I=7

48ML

2+2M

L

4

!

"#

$

%&2

+3M3L

4

!

"#

$

%&2

I=94

48ML

2

= I

+rF2mg

rFg rF3mg = I

L

42mg

L

4mg

3L

43mg =

94

48ML

2#$%

&'(

=96

94

g

L=

48

47

g

L

8.a. The satellites orbit is maintained by the force of earths gravity, acting as a centripetal force.

Fgravity = Fcentripetal

GMm

r2

=mv

2

r

v =GM

r=

GM

rearth + altitude( )

v = 4.9e3m / s

b. The radial acceleration can be calculated in one of two ways:ac=

v2

r=G

M

r2

ac=1.49m / s

2

c. The period of the satellitethe time it takes to complete one orbitcan be calculated as follows:T=

2

=

2r

v

T=2(6.38e6+1.00e7)

4.9e3= 2.1e4s = 5.8hrs

d. Total mechanical energy is the sum of kinetic and potential energies:E

total= K+U

Etotal

=

1

2mv

2+G

Mm

r

Etotal

=

1

2617(4.9e3)

2+(6.672e11)

(5.98e24)(617)

6.38e6+1e7( )

Etotal

= 7.6e9J

The shortcut for solving this problem is remembering that, for bound orbits, Etotal

=

1

2U= G

Mm

r.

e. The satellite has potential energy at the surface of the earth, and needs additional energy (kinetic) toachieve the total energy calculated in part (d).


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Etotal

= K+U

7.6e9J=1

2mv

esc

2+G

Mm

r

vesc =

2

m 7.6e9J+GMm

r

"

#$

%

&'

vesc=

2

6177.6e9J+6.672e11

(5.98e24)(617)

6.38e6

"

#$

%

&'

vesc=1.12e4m / s

f. The center-of-mass can be calculated usingx

cm=

xearth

mearth

+ xsatellite

msatellite

mearth

+msatellite

Considering center of earth as origin:

xcm=

0(5.98e24)+1.638e7)(617)

5.98e24+617=1.69e15m

This is obviouslyextremelyclose to the center of the earth.

9.a. When the 1.00kg platform is placed on top of the

spring, the spring is compressed until the force ofgravity pulling the platform down is balanced by theforce of the spring pushing upward. At that point,

Fnet=ma = 0

+Fspring +Fgravity = 0

kx mg = 0

x =mg

k=

(1.0kg)(9.8m / s2)

500N /m= 0.020m

b. Calculating period:T= 2

m

k

T= 21.00kg

500N /m= 0.28s

k=500N/m

m=1.00kg

Fspring

Fgravity


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c. The energy of the platform is based on the stored potential energy when the spring has beencompressed 0.10m. At the equilibrium position, that elastic potential energy has been converted toboth gravitational and kinetic energy, which allows us to calculate velocity.

Ebottom = Eequilibrium

Uelastic =Ugravitational +K

1

2kx2 =mgh+

1

2mv2

1

2(500)(0.10)

2= (1)(9.8)(0.10)+

1

2(1)v2

v =1.7m / s

d. Using the same process as beforeT= 2

m

k

T= 21.50kg

500N /m= 0.34s

e. The amplitude of the new system is measured as the maximum displacement from the equilibriumposition. The new equilibrium position (relative to the unstretched spring, is

Fspring = Fgravity

kx =mg

x =mg

k=

(1.5)(9.8)

500= 0.0298m = 0.030m

We also need to determine the maximum displacement of the spring, which can be calculated basedon an energy analysis, where the kinetic energy just after the collision and the changing gravitationalpotential energy are converted to elastic potential energy.

Ill begin by calculating the velocity of the platform-blob just after the collision, using Conservationof Momentum:

pblob + pplatform = pblob+platform

mblobvblob +0 = (mblob +mplatform) !v

Looks like I need to get the velocity of theblog just before it hits, so Ill do that usingenergy:

Ui +

Ki =

Uf +

Kf

mgh =1

2mv2

v = 2gh = 2 9.8 0.30m = 2.4m / s

Now:

k=500N/m

m=1.00kg

h=30.0cm

m=0.50kg

h=0.030

x0

A


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2011 Richard White www.crashwhite.com

pblob + pplatform = pblob+platform

mblobvblob +0 = (mblob +mplatform) !v

(0.5kg)(2.4m / s)= (0.5+1.0) !v

!v=

0.8m / s

For simplicitys sake, take the height of impact to be h=0.030m for my Energy analysis (althoughyou can use any position you want, as long as you keep track ofchangesin potential energy):

Einitial = EfinalUg +K+Uspring =Ug +K+Uspring

mghi +1

2mvi

2+

1

2kxi

2=mghf +

1

2mvf

2+

1

2kxf

2

(1.5)(9.8)(0.03)+1

2(1.5)(2.4)2 +

1

2(500)(0.03)2 = (1.5)(9.8)(A))+0+

1

2(500)A2

Solve quadratic equation to get

A = {0.174m,0.115m}

Because Ive solved this problem with a negative in my height, the correct Amplitude solution is the0.174 meters displacement. Plug this value into the equation to confirm that it correctly satisfies ouranalysis.

f. The period for the oscillating system will be the sameas it was on Earth. According to the formula forperiod of a mass-spring system, the only two factors that determine the period are the mass mandthe spring constant k, and those two values are the same on Mars as they were on Earth.

g. The amplitude for the oscillating system will be less thanit was on Earth. The weaker gravity field willresults in a smaller velocity for the class blog just before it hits the platform, and the change inpotential energyU=mgh will be less, also due to a smaller g value. With less initial energy in

the system, there will be less elastic potential energy stored in the spring, and thus a smaller

amplitude when Uspring is at a maximum.

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Practice Test 7 Static Gravitation Shm

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