Question

A potter's wheel with rotational inertia 7.5 kg m2 is spinning freely at 16.0 rpm. The potter drops a 2.65 kg lump of clay on the wheel, where it sticks 46.0 cm from the rotation axis. what's the wheel's subsequent angular speed

Answer 1

Answer:

1.56 rad/s

Explanation:

$\omega_1$ = 16 revolution/minute = 16 * 2π (rad/rev) * (1/60) (min/sec) = 1.68 rad/s

46 cm = 0.46 m

Treating the lump of clay as a point mass 46 cm from the rotational axis, the new rotational inertia of the system would be

$I_2 = I_1 + mr^2 = 7.5 + 2.65*0.46^2 = 8.06 kgm^2$

According the the law of momentum conservation, the product of rotational inertia and angular speed must stay the same:

$I_1\omega_1 = I_2\omega_2$

$\omega_2 = \omega_1\frac{I_1}{I_2} = 1.68 * \frac{7.5}{8.06} = 1.56 rad/s$