A hamster sits 0.10 m from the center of a lazy Susan of negligible mass. The wheel has an angular velocity of 1.0 rev/s. How will the angular velocity of the lazy Susan change if the hamster walks to 0.30 m from the center of rotation? Assume zero friction and no external torque.
a) It will speed up to 2.0 rev/s.
b) It will speed up to 9.0 rev/s.
c) It will slow to 0.01 rev/s.
d) It will slow to 0.02 rev/s.
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To determine how the angular velocity of the lazy Susan will change when the hamster walks to a new position, we can use the principle of conservation of angular momentum.
The angular momentum of the system remains constant when there is no external torque acting on it. The angular momentum can be calculated as the product of the moment of inertia and the angular velocity:
L = I * ω
As the hamster moves away from the center of rotation, the moment of inertia of the system will change. However, since the question states that the lazy Susan has negligible mass, we can assume that the moment of inertia remains constant.
Therefore, we can write:
I1 * ω1 = I2 * ω2
where I1 is the initial moment of inertia, ω1 is the initial angular velocity, I2 is the final moment of inertia, and ω2 is the final angular velocity.
Since the moment of inertia remains constant, we can simplify the equation to:
ω1 = I2 * ω2 / I1
Let's calculate the initial and final moment of inertia:
I1 = m * r1^2
I2 = m * r2^2
where m is the mass and r1 and r2 are the initial and final distances from the center of rotation, respectively.
Given:
r1 = 0.10 m
r2 = 0.30 m
ω1 = 1.0 rev/s
Substituting these values into the equation, we get:
1.0 = (m * r2^2 * ω2) / (m * r1^2)
Simplifying further:
1.0 = (r2^2 * ω2) / r1^2
Multiplying through by r1^2:
r1^2 = r2^2 * ω2
Substituting the values:
(0.10 m)^2 = (0.30 m)^2 * ω2
0.01 m^2 = 0.09 m^2 * ω2
Simplifying:
0.01 = 0.09 * ω2
ω2 = 0.01 / 0.09
ω2 = 0.11 rev/s
Therefore, the angular velocity of the lazy Susan will slow down to 0.11 rev/s.
So, the correct answer is:
d) It will slow to 0.02 rev/s.