Question

A bowling ball of uniform density is projected along a horizontal with a velocity $v_0$ so that it initially slides without rolling. The ball has mass $m$ and coefficient of static friction ${\mu }_d$ with the floor. Ignore air-friction. Let $t$ be the time at which the ball begins to roll without sliding and $v$ be the velocity of the ball when this happens. Then:

• A. $t=\frac{2v_0}{7{\mu }_{d}g}$
• B. $t=\frac{1}{7}\frac{v_0}{{\mu }_{s}g}$
• C. $v=\frac{5}{7}{v}_0$
• D. $v=\frac{3}{7}{v}_0$

Answer 1

Answer: (A), (C)

The correct options are
A $t=\frac{2v_0}{7{\mu }_{d}g}$
C $v=\frac{5}{7}{v}_0$

Friction not only causes v to decrease but produces a torque which gives an angular acceleration $\alpha$ causing $\omega$ to increase.
$F_{fr}=\mu N=\mu mg$
$\tau =F_{fr}R=I\alpha$
$\Rightarrow \mu mgR=\frac{2}{5}m{R}^2\alpha$
$\Rightarrow \alpha =\frac{5\mu g}{2R}$
Hence, from equation of translational motion, $v=u+at$ we have $v=v_0+at=v_0-\mu gt$ as $u=v_0$ and $a=-\mu g$
From equation of rotational motion, $\omega ={\omega }_0+\alpha t=0+\frac{5\mu g}{2R}t=\frac{5\mu g}{2R}t$ as ${\omega }_0=0$
In case of rolling without sliding, $v=\omega R$
$\Rightarrow (v_0-\mu gt)=R\left(\frac{5\mu g}{2R}\right)t$
$v_0=\mu gt+\frac{5}{2}\mu gt=\frac{7}{2}\mu gt$
$t=\frac{2v_0}{7\mu g}$
Substituting the value of t in $v=v_0-\mu gt$
we get $v=v_0-\mu g\frac{2v_0}{7\mu g}=\frac{5}{7}{v}_0$