Download Solutions to HW 2

Solutions to HW 5
PY 105
1)
While on a train, you are holding a string with a ball attached to it. At first, the train’s
velocity is constant and the string is vertical. When the train changes direction, while going
around a curve with a radius of 200 m, you measure the angle between the string and the
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vertical to be 34.0 degrees. The ball has a mass of 1.00 kg, and we will use g = 10 m/s .
(a) The speed of the train is constant during this time. What is that speed?
A quick free body diagram shows that the horizontal component of the tension is the force
responsible for the centripetal acceleration.
Hence writing out Newton's second law in the x and y directions yields:
in the x:
in the y:
which results in
Dividing the equation for x by the equation for y eliminates the tension and replacing a c
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with v /r results in:
and solving for the velocity gives
or, for this specific problem, 36.7 m/s
(b) What is the tension in the string while the train is going around the curve?
Returning to the equations in x and y, we can now solve for T. I choose the y direction
equation because it is slightly simpler:
or, for this specific problem, about 12 newtons.
2)
Mary is whirling around on a merry-go-round at the playground that her friend Caitlin is
gradually spinning faster and faster. Determine the maximum speed Mary can go, if the
coefficient of static friction between her shoes and the floor is 0.450. Assume that Mary is
not hanging on, but is simply balancing on her feet at a distance of 4.00 m from the center
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of the merry-go-round. Use g = 10 m/s .
(a) Mary's maximum speed is?
The horizontal force of friction must be the force providing the centripetal acceleration.
Writing out newton's second law for the x and y direction results in:
in the x:
in the y:
which gives
This results in
and solving for the velocity yields
or, for this specific problem, v = 3 * sqrt(2) = 4.24 m/s
(b) Which, if any, of the following would help Mary stay on the merry-go-round without
sliding? Select all that apply.
correct
incorrect
incorrect
Moving a little closer to the center of the merry-go-round.
Moving a little closer to the outer edge of the merry-go-round.
Wearing a heavy backpack that increases her mass.
In order to understand the answer (which was incorrect in webassign) it is best to examine
the equation
. Here the left side represents the maximum force that friction can
apply to hold Mary in a circular path. The right hand side represents the necessary
centripetal acceleration. Thus to help Mary NOT slip, the right hand side should decrease.
It is important here to realize that for an object rotating, the farther from the central axis,
the faster the tangential velocity. This is because the object must travel a larger distance in
the same amount of time. Hence, even though the equation is inversely proportional to the
radius, moving toward the center of rotation reduces the necessary centripetal acceleration
and, therefore, reduces the necessary frictional force required by the shoes.
3)
As shown in the figure, a disk of mass m = 500 grams is placed on a rotating turntable so
that the disk spins with the turntable at a constant radius of 40.0 cm from the center. The
disk has a constant speed. The disk is tied to a cylinder, which has a mass of 3m, by a
string that passes through a hole in the center of the turntable. As the disk and the
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turntable spin, the cylinder maintains its position vertically. Use g = 10 m/s .
(a) Determine the value of the tension in the string. Assume that there is no friction
between the disk and the turntable.
Since the hanging cylinder is not moving, newton's second law can be written as:
which, by solving for T gives T = 3mg, or for this specific problem, T = 15 N
(b) Determine the speed of the disk.
Now by examining the block that is spinning it is clear that the centripetal acceleration is
created by the tension force acting on the rotating disk. Hence
and from part (a) T is a known quantity. Thus substituting and solving for v
gives
or for this specific problem, 2*sqrt(3) = 3.46 m/s
4)
A cube with a mass of 400 grams is placed in a funnel that is rotating around the vertical
axis, as shown above. There is no friction between the cube and the funnel, but the funnel
is rotating at just the right speed needed to keep the cube rotating with the funnel. The
cube travels in a circular path of radius r = 30.0 cm, and the angle between the vertical and
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the wall of the funnel is 32.0 degrees. Use g = 10 m/s .
(a) Which force, forces, or force components give rise to the cube's centripetal acceleration
in this situation? Select all that apply.
incorrect The full force of gravity.
incorrect The horizontal component of the force of gravity.
incorrect The vertical component of the force of gravity.
incorrect The full normal force.
correct
The horizontal component of the normal force.
incorrect The vertical component of the normal force.
incorrect The mysterious centripetal force, which comes from neither the force of gravity
nor
the normal force, but exists simply because the cube is moving in a
circular path.
Since the acceleration is horizontal (with respect to the picture shown above) the force that
causes the acceleration must also be in the horizontal direction. The only force that acts in
this direction is the horizontal component of the normal force and therefore this is the only
force that may give rise to the centripetal acceleration.
(b) Determine the value of the normal force acting on the cube.
I will choose my coordinate system such that x is in the horizontal direction and y is in the
vertical direction. A free body diagram follows:
Thus setting up newtons laws in the x and y directions gives:
in the x:
in the y:
Using the y equation to solve for the normal force it is found that
N = mg/sin(theta) or for this specific problem: 4 sin (32) = 2.11 Newtons
(c) Determine the speed of the cube.
Using the above equation for x and substituting in the value for the normal force yields:
. By solving for velocity
and, for this specific problem, 2.2 m/s
5)
Matt, with a mass of 23.0 kg, is swinging on a tire swing (a tire attached to a rope hanging
from a tree). The tire itself has a mass of 5.0 kg, and the rope has a length of 4.00 m. Use
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g = 10 m/s . Matt’s speed is 3.00 m/s when the tire reaches the bottom of its circular arc,
what is the tension in the rope at that point?
At the bottom of his path, tension is pulling upward while gravity pulls down on both Matt
and the tire. In addition, because the tire is undergoing circular motion, the tire must be
accelerating upward with ac. By applying Newton's second law:
or
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Therefore, T = 28(10) + (28)(3 )/4 = 280 + 63 = 343 newtons