A thin, 108 g disk with a diameter of 8.5 cm rotates about an axis through its center with 0.25 J of kinetic energy. What is the speed of a point on the rim?
Question
A thin, 108 g disk with a diameter of 8.5 cm rotates about an axis through its center with 0.25 J of kinetic energy. What is the speed of a point on the rim?
Solution
Sure, here are the steps to solve this problem:
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First, we need to calculate the moment of inertia (I) of the disk. The formula for the moment of inertia of a disk rotating about an axis through its center is I = 0.5mr^2, where m is the mass and r is the radius.
Given that the mass m = 108 g = 0.108 kg (since 1 g = 0.001 kg) and the diameter is 8.5 cm = 0.085 m (since 1 cm = 0.01 m), the radius r = 0.085 m / 2 = 0.0425 m.
So, I = 0.5 * 0.108 kg * (0.0425 m)^2 = 0.000097 kg*m^2.
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Next, we use the formula for the kinetic energy of a rotating object, which is KE = 0.5Iw^2, where w is the angular velocity. We can rearrange this formula to solve for w: w = sqrt(2*KE/I).
Given that KE = 0.25 J, we find that w = sqrt(20.25 J / 0.000097 kgm^2) = 72.3 rad/s.
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Finally, we find the speed v of a point on the rim using the formula v = w*r, which gives v = 72.3 rad/s * 0.0425 m = 3.07 m/s.
So, the speed of a point on the rim is approximately 3.07 m/s.
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