#$&*
course Phy 201
12/6 1 pm Below is a full test that I completed this morning since we didn't have class. Any critiques/comments are appreciated.
Problem Number 1
What would be the orbital KE of a satellite of mass 480 kg in circular orbit about a planet of mass 88 * 10^24 kg, orbiting at a distance of 15200 km from the center of the planet?
By how much would orbital PE change as the satellite moved from this orbit to a circular orbit of radius 16720 km?
G = 6.67 * 10^-11 N m^2 / kg^2
PE=-G *Mm/r KE=.5PE
PE=-6.67*10^-11 N m^2/kg^2*(88*10^24 kg)(480kg)/15200000m
PE=-1.85 x10^11 J
KE=9.27 x10^10 J
Change in PE=-1.64 x10^10
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Problem Number 2
A simple pendulum has a length of 2.45 meters and a mass of .34 kg. It is given a KE of .187 Joules at a point .1641 meters from equilibrium. What will be its maximum displacement from equilibrium?
Looking for x +.1641 m
KE=1/2 mv^2
.187 J=1/2(.34 kg)(v^2)
.187 J/.17 kg=v^2
1.1 m^2/s^2=v^2
v=1.049 m/s
K=(1.049m/s)^2*.34=.374Nm
F=-kx---I am not sure where to go with this problem.
The pendulum also has KE, equal to 1/2 k x^2. This is a pendulum, so k = m g / L.
Add the KE and the PE to get the total energy.
At max displacement KE will be zero, to PE will be equal to the total energy.
Thus if max displacement is A, the equation 1/2 k A^2 = total energy can be solved for A.
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Problem Number 3
A simple harmonic oscillator has a restoring force of 350 N/m and a mass of .39 kg. It is given a KE of .1794 Joules at a point .1862 meters from equilibrium. What will be its maximum displacement from equilibrium?
K=350N/m m=.39 kg
PE=1/2kx^2
PE=1/2 (350Nm)(.1862 m)^2=6.07 Nm
This problem is similar to one of the problems I’ve already asked about. Since energy is not “lost”. Would PE + KE=1/2kA^2?
This is exactly correct. See also my note on the preceding question, where the only additional task would be to find k.
If so, then
6.25 J=1/2 (350Nm)(A^2)
A=sqrt(2*6.25 J/350 Nm)
A=0.189 m
I used the question previously to guide me through, but I’m not convinced it is 100% correct.
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Problem Number 4
A simple harmonic oscillator is subjected to a net restoring force F = - 400 N/m * x at displacement x from equilibrium. It is observed to undergo simple harmonic motion with a frequency of 1.3 cycles / second. What is its mass?
2 pi rad* 1.3 cycles/second=8.17 rad/s
8.17 rad/s= sqrt(-400Nm/m), square both sides
66.7 rad^2/s^2=(-400Nm/mass)
mass=-400Nm/66.7 rad^2/s^2=-6 kg. CAN Mass be negative? I wouldn’t think so….so mass equals 6 kg
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F = - k x. The negative is in the equation. k itself isn't negative.
Except for the - sign your solution is correct.
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Problem Number 5
A person is being rotated in a horizontal circle of radius 9 meters. If the person feels a centripetal force of 5.1 times her own weight, how fast is she traveling?
5.1(mg)=m v^2/r “m” cancels out
5.1g=v^2/r
v=sqrt(5.1g/r)
v=sqrt (5.1*9.8 m/s^2/9 meters)
v=sqrt(49.98 m/s^2/9 meters)
v=21.1 m/s
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Problem Number 6
A roller coaster is at the top of a circular loop of radius 20 meters. The roller coaster is turned upside down. How fast must it be moving for a passenger sitting on a scale to register her usual weight?
r=20 meters
If it registers her weight, then centripetal force equals 2mg
2mg=m v^2/r
2g=v^2/r
v=sqrt (2g/r)
v=sqrt((2*9.8 m/s^2)/20 m).
v=.989 m/s (should this be rad/s?)
2 g = v^2 / r has solution v = sqrt( 2 g * r), not sqrt( 2 g / r). The units then work out to m/s.
I'm glad you're paying attention to the units.
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Problem Number 7
Explain why the work required to stretch a spring or other elastic object with a linear restoring force, of form F = - kx, from its equilibrium position to displacement x is `dW = .5 k x^2, and why we hence say that this is the elastic potential energy of the object in this position.
The forces that do the work are linear from 0 to k*x; therefore averages to 1/2k*x. The force is done through the distance x so 1/2k*x*x=1/2kx^2. When released, it will return to equilibrium with forces equal and opposite to those that brought it to that position and be regained in the form of KE.
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Problem Number 8
If a mass of 5 kg moving at 4 m/s collides with a mass of 8 kg moving at -9 m/s, and the two masses are 'stuck together' after collision, then what is their common velocity after collision? Is this collision possible for the system consisting of the two masses without the conversion of some internal source of potential energy?
M1=5 kg v1=4 m/s m2=8 kg v2=-9 m/s
p1=5 kg*4 m/s=20 kg m/s p2= 8 kg*-9 m/s=-72 kg m/s
20 kg m/s + (-72 kg m/s)=-52 kg m/s
Total mass is 13 kg: -52 kg m/s/13 kg= -4 m/s (common velocity)
For the second part of the question, I am going to answer that yes it is possible for this collision, but I am not sure why.
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To answer the second question you need to determine whether the kinetic energy after collision is less than or equal to that before collision. If the system gains kinetic energy, it has to be converted from somewhere within the system.
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Problem Number 9
An Atwood's machine consists of a disk of radius 19 cm and mass 2.5 kg, with masses of .5 kg and .65 kg suspended from a cord over the disk. If the cord does not slip on the disk, what is the total kinetic energy of this system when the masses are moving at 25 m/s?
.Disk= r =19 cm, .19 m mass= 2.5 kg
2 masses suspended .5 kg and .65 kg. Moving at 25 m/s Total KE
First find moment of inertia
disk=1/2MR^2
Idisk=1/2 2.5 kg(.19 m^2)
Idisk=.045 kg m^2
Mass 1=
I was going to solve the above problem in that manner, but not having R for the second 2 masses, made me reconsider….
KE=1/2 m v^2
KE=1/2 (3.65 kg)(25 m/s)^2
KE=1.825 kg(625 m^2/s^2)
KE=1140.6 J
You can't add the mass of the pulley to the suspended masses. Most of the pulley mass is moving more slowly than the suspended masses.
The KE of the pulley is 1/2 I omega^2. You have the radius of the pulley so you can figure out its angular velocity.
Then add this to the 1/2 m v^2 of the two suspended masses.
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Problem Number 10
A pendulum of mass 1.2 kg and length 2.6 meters is initially displaced .7539999 cm in the horizontal direction from its equilibrium position. The pendulum is released from rest at this position.
• How much work does gravity therefore do on the pendulum during its fall to the equilibrium position, and how much work does the pendulum do against gravity?
• What therefore will be the change in the kinetic energy of the pendulum as it falls?
• Using energy considerations determine the velocity of pendulum at its equilibrium position.
.m=1.2 kg L=2.6 m `dx=.754 cm (.00754 m)
mg=11.76 N
I know that W=Fparallel through the distance. How do you find the parallel force of the pendulum?
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The motion is almost all in the horizontal direction. Let that be the x direction, with y vertical.
You would use the horizontal component of the tension.
The angle is small so the tension is very nearly equal to the weight of the pendulum.
The magnitude of the cosine of the angle of the tension force is x / L, where x is the displacement and L the length.
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Problem Number 11
If a simple pendulum of length 2.09 meters is subjected to a restoring force of 5.5 Newtons when displaced .1337 meters from equilibrium, what is the mass of the pendulum? What will be its KE and its PE at equilibrium and halfway to equilibrium if it is released from rest at a point .1337 meters from equilibrium?
Mass of pendulum;
k=mg/L
5.5N=m (9.8 m/s^2)/2.09 m
You've set the force equal to k.
k = m g / L.
The restoring force has magnitude | F | = | k x |.
11.5 Nm=m(9.8 m/s^2) divide by 9.8 m/s^2
1.17 kg=mass
as a guideline, I believe the mass is more like 8 kg, give or take a kg or so
PE=1/2kx^2
at rest PE= ˝ 5.5N*(.1337m)^2
k isn't 5.5 N
PE=.0495 J
KE at rest=0
PE at halfway=1/25.5N (.067 m)^2=.012 J
Then KE=.0495 J-.012 J=.0375J
At equilibrium
PE=0
KE=.0495 J
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You're on the right track overall, but you've confused F with k. Correct that and it will work out.
Again for a guideline, I think k is in the neighborhood of 40 N / m.
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Problem Number 12
Sketch a force diagram depicting weight and normal force for a glider gliding down an inclined air track. Indicate an x axis parallel to the incline and a y axis perpendicularto the x axis. Sketch the force vectors to scale and indicate their x and y components. Also sketch the net force vector, in the appropriate direction and also to scale.
Sketch a diagram to show the forces that would result if a horizontal force equal to half the weight of the glider was exerted in a direction that would tend to push the glider up the incline. Indicate the x and y components of all these vectors. Sketch all vectors and all components to scale.
We know that for an air cart gliding along an incline, the force that accelerates it is equal to W sin(`theta), where W is the weight of the cart and `theta the angle of the incline with horizontal.
If the force on the glider was W, what would be its acceleration?
a=W/m
W = m g, so a = W / m = m g / m = g.
If the acceleration of a mass is proportional to the net force acting on it, then
what should be the acceleration of the glider under the influence of a net force of .04 W?
a=.04W/m
Using W = m g you get a = .04 g.
what should be the acceleration of the glider under the influence of net force W sin(`theta)?
a=Wsin(theta)/m
Using W = m g you get a = g sin(theta).
What should be the acceleration of the glider for `theta = 3 degrees?
a=Wsin 3 degrees
a=Wsin 3 degrees / m
What is the slope of the incline corresponding to this angle?
Mgx/mg
Over a displacement `ds along the incline the rise would be `ds sin(theta), the run would be `ds cos(theta) and the slope would be rise / run = `ds sin(theta) / (`ds cos(theta) ) = sin(theta) / cos(theta) (also equal to tan(theta)). So the slope corresponding to a 3 deg angle would be sin(3 deg) / cos(3 deg), a little over .05 when you calculate it.
Explain why, for small angles, acceleration is proportional to slope.
Acceleration is proportional to slope because the mgy component is approximately equal to mg and it is the force acting on the cart.
One way to see it:
slope = sin(theta) / cos(theta) and for small angles cos(theta) is close to 1. So slope is close to sin(theta). Since a = g sin(theta), we have the very close approximation (provided the angle is small) that a = g * slope.
Another way:
The weight vector forms a right triangle with legs equal to its x and y components.
A displacement along the slope forms a right triangle with legs equal to its x and y components.
These triangles both include the 3 degree angle. The hypotenuse and the longer leg are very nearly the same in each triangle.
The parallel component of the weight vector is in the same proportion to the weight as the 'rise' leg to the displacement along the incline, which in turn is very nearly equal to the 'run' of that triangle.
So the parallel component of the weight vector is very equal to slope * weight.
It follows that the acceleration is very nearly equal to slope * weight / m = slope * m g / m = slope * g.
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