A solid sphere is set into motion on a rough horizontal surface with a linear speed $v$ in the forward direction and an angular speed $\frac{V}{R}$ in the anticlockwise direction as shown in figure. Find the linear speed of the sphere when it stops rotating and $\omega=\frac{V}{R}$ .

Question

A solid sphere is set into motion on a rough horizontal surface with a linear speed

v

in the forward direction and an angular speed

\frac{V}{R}

in the anticlockwise direction as shown in figure. Find the linear speed of the sphere when it stops rotating and

\omega=\frac{V}{R}

.

A solid sphere is set into motion on a rough horizontal surface with a linear speed v in the forward direction and an

Quantrex Academy · Accepted Answer

Correct answer: A. (3 v)/(5)

If we take moment at A , then extemal torque will be zero. Therefore, the initial angular momentum = the angular momentum after rotation stop (i.e. only linear velocity exist) aligned & M v × R-((2)/(5)) I ω & =M v_0 × R & M v R-(2)/(5) M R^2 × (V)/(R) & =M v_0 R & v_0 & =(3 v)/(5) aligned

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