Question

This is may mastering physics homework and I just need help solving the question.

Answer 1

Answer:

Angular speed of the disc after he fold his hands is given as

$\omega_f = 27.5 rpm$

Explanation:

As we know that the moment of inertia of the solid cylinder is given as

$I_1 = \frac{m_1r_1^2}{2}$

so we have

$I_1 = \frac{(950/9.81)(0.20)^2}{2}$

$I_1 = 1.94 kg m^2$

now moment of inertia of two dumbbell in his hand is given as

$I_2 = 2(m_2 r_2^2)$

$I_2 = 2(4)(0.85)^2$

$I_2 = 5.78 kg m^2$

Now moment of inertia of the disc is given as

$I_3 = \frac{1}{2}MR^2$

$I_3 = \frac{1}{2}(1500/9.81)(2^2)$

$I_3 = 305.8 kg m^2$

Now we can use angular momentum conservation as there is no external torque on it

$(I_1 + I_2 + I_3) \omega_i = (I_1 + I_3)\omega_f$

$(1.94 + 5.78 + 305.8) 27 = (1.94 + 305.8) \omega_f$

$\omega_f = 27.5 rpm$