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System Of Particles And Rotational Motion

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 ω = $\frac{v}{R}$

$\frac{3 v}{5}$

$\frac{2 v}{5}$

$\frac{4 v}{3}$

$\frac{7 v}{3}$

Solution

$\frac{3 v}{5}$

If we take moment at A, then external torque will be zero.

Therefore,

the initial angular momentum= the angular momentum after rotation stop

(i.e. only linear velocity exist )

(i) Initial angular momentum about point A

L = m ( v × R ) - Iω

⇒ Mv × R - $(\frac{2}{5})$ Iω = M v₀ × R

⇒ MvR - $\frac{2}{5}$ MR² × $\frac{v}{R}$ = Mv₀ R

⇒ v₀ = $\frac{3 v}{5}$