A uniform horizontal circular disc of mass M and radius R is rotating freely about a vertical axis passing through its center at an angular speed of 40 rpm . Two small blocks, each of mass (M)/(4) , are initially fixed

Question

Quantrex Academy · Accepted Answer

Correct answer: C. 64 rpm

The initial moment of inertia of the system is the sum of the moment of inertia of the disc and the two blocks at the edge.

I_1 = \dfrac{1}{2}MR^2 + 2\left(\dfrac{M}{4}R^2\right) = \dfrac{1}{2}MR^2 + \dfrac{1}{2}MR^2 = MR^2

When the blocks are moved to a distance of

\dfrac{R}{2}

from the center, the new moment of inertia is:

I_2 = \dfrac{1}{2}MR^2 + 2\left(\dfrac{M}{4}\left(\dfrac{R}{2}\right)^2\right) = \dfrac{1}{2}MR^2 + \dfrac{1}{8}MR^2 = \dfrac{5}{8}MR^2

Since no external torque acts on the system, angular momentum is conserved:

I_1\omega_1 = I_2\omega_2

Substituting the given values:

MR^2 \times 40 = \dfrac{5}{8}MR^2 \times \omega_2

\omega_2 = 40 \times \dfrac{8}{5} = 64\text{ rpm}

Answer:

64\text{ rpm}

A uniform horizontal circular disc of mass M and radius R is...

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