Question

d

Since the spokes of a bicycle wheel have much less mass than the rim, the wheel can be modeled as a cylindrical hoop of mass 1.90 kg and radius 0.6 m. This wheel rotates without friction about a vertical axis perpendicular to the disk and through its center. Standing on the rim is a kitten of mass 1.00 kg (the kitten can be modeled as a point mass). The wheel (with the kitten standing on it) rotates with an angular velocity of 0.600 revolutions per second.

a) What is the moment of inertia of the wheel/kitten system?

b) The kitten then scampers from its initial position inward on the wheel until it is standing at the center. What is the new moment of inertia of the wheel/kitten system?

c) What is the angular velocity of the system when the kitten reaches the center?

d) What is the ratio of the system

Answer 1

This question requires conservation of angular momentum, rotational kinetic energy, and knowing that the moment of inertia for a thin hoop and a point mass are both mr².

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Part (a)

Step 1) Since the system of the kitten and wheel is composed of two objects, the total moment of inertia I_T of them will be:

with I_w and I_k being the moments of inertia for the wheel and kitten respectively.

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Step 2) Since the moment of inertia for a thin hoop and a point mass around an axis are both mr², use that to fill in for the moments, and simplify slightly:

The same radius r is used for the wheel and kitten because the kitten is sitting on the rim, and thus has the same distance from the axis of rotation as the rest of the wheel.

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Step 3) Fill in 1.90 kg for the mass of the wheel m_w, 1.00 kg for the mass of the kitten m_k, and 0.6 m for the radius r:

So the total moment of inertia for the wheel/kitten combination is 1.044kg*m².

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Part (b)

Step 1) To find the moment of inertia when the kitten runs to the axis of rotation, use the equation for the total moment found before, but use different radii for the wheel and kitten:

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Step 2) Then fill in the same information as before, except using 0 for the radius for the kitten (since it's sitting on the axis):

So the moment of inertia when the kitten is sitting at the center is 0.684 kg*m².

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Part (c)

Step 1) To find the new angular velocity after the kitten runs to the center, start with conservation of angular momentum:

with I_Ti being the initial total moment of inertia, I_Tf being the final total moment of inertia, being the initial angular velocity, and being the final angular velocity.

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Step 2) Solve for the final angular velocity:

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Step 3) Fill in 1.044kg*m² for the initial moment of inertia found in part (a), 0.684 kg*m² for the final moment of inertia found in part (b), and 0.600 rev/s for the initial angular velocity:

So the final angular velocity after the kitten runs to the center is about 1.53 revolutions per second.

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Part (d)

Step 1) To find the ratio R of the system's final kinetic energy Kf to that of it's initial energy K_i set up the equation for it:

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Step 2) Use that the rotational kinetic energy of a system is (1/2)I, and simplify slightly:

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Step 3) Fill in 0.684 kg*m² for the final moment of inertia, 0.044 kg*m² for the initial moment of inertia, 1.526 rev/s for the final angular velocity, and 0.600 rev/s for the initial angular velocity:

So the system contains 4.24 times as much rotational kinetic energy at the end as it did at the start.

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Part (e)

Energy is not conserved here because the kitten had to do work in order to run to the center. It's that energy that the kitten used that got converted into the additional rotational energy of the system.