Question

A bowling ball is thrown down the alley in such a way that it slides with a speed $v_0$ initially without rolling. What will it it's speed be when (and if) it starts pure rolling? The transition from pure sliding to pure rolling is gradual, so that both sliding and rolling takes place during this interval.

• A. $\frac{5}{7}{v}_0$
• B. $\frac{2}{3}{v}_0$
• C. $\frac{5}{12}{v}_0$
• D. None of these

Answer 1

The correct option is A

$\frac{5}{7}{v}_0$

1. Force/Torque Method:

The forces acting on the ball are as shown in the figure. Let the acceleration of centre of mass be acm and its angular acceleration about the centre of mass axis be α as shown in the figure.

The following figures represents the forces, accelerations and velocities of the ball in three different positions.

Force equation are: N - mg = 0(In the vertical direction) - fk = macm (In the horizontal direction)

⇒acm = fk / m . . . . . . (1)

And from $tau=I_{cm}\alpha \Rightarrow f_{k}r=\left(\frac{2}{5}m{r}^2\right)\alpha \Rightarrow \alpha =\left(\frac{f_{k}r}{\frac{2}{5}m{r}^2}\right)$. . . . . . (2)

Let the velocity of the center of mass be $v_{cm}$and angular velocity about the center of mass axis be w when rolling starts

$v_{cm}=v_0+\left(\frac{-f_k}{m}\right)t$ ....(3)

using v=u+at

and $\omega =0+\left(\frac{f_{k}r}{\frac{2}{5}m{r}^2}\right)t$ ....(4)

$\omega ={\omega }_0+at$

Also $v_{cm}=r\omega$

solving equations (3), (4) and (5) for $v_{cm},v_{cm}=v_0$

(Calculate the value of $f_{k}t$ from equation (3) and substitute it in equation (4). Substitute $v_{cm}/rfor\omega )$.

2. Work - Energy Theorem:

Apply the work energy theorem for the translation and the rotational component of the motion separately.

$\left(i\right)\sum W_{force}=\Delta K_{translational}$

$\left(ii\right)\sum W_{torque}=\Delta K_{relational}$

Translate the normal force and the frictional force at the centre of mass and apply an equivalent torque about the centre of mas to resolve the motion of the ball into translation and rotational components as shown in the figure.

Mathematically,

$W_{fk}+W_N+W_{mg}=k_2-K_1$

translational

$\Rightarrow -f_{k}x+0+0=\frac{1}{2}m{v}^2-\frac{1}{2}m{v}_0^2$

$\Rightarrow -f_k{v}_{av}.t=\frac{1}{2}{v}^2-v_0^2\because x=v_{av}.t$

$\Rightarrow -f_k\left(\frac{v+v_0}{2}\right).t=\frac{1}{2}mv-v_{0}v+v_0$ ...(1)

(in case of constant acceleration only $v_{av}=\frac{v_i+v_r}{2}$)andfor the rotational component of motion,

$f_{k}r.\theta =\frac{1}{2}{I}_{cm}{\omega }^2-0\Rightarrow f_{k}r{\omega }_{av}.t=\frac{1}{2}{I}_{cm}{\omega }^2\Rightarrow f_{k}r\left(\frac{0+\omega }{2}\right)t=\frac{1}{2}\frac{2}{5}m{r}^2.{\omega }^2$

[in case of constant angular acceleration ${\omega }_{av}=\frac{{\omega }_0+{\omega }_r}{2}$]$\Rightarrow f_{k}r\frac{\omega }{2}t=\frac{1}{2}\frac{2}{5}m{r}^2.{\omega }^2$ ...(2)

From equation (1) and (2),

$\Rightarrow \frac{-f_k.\frac{v+v_0}{.}t{f}_{k}r.\frac{\omega }{2}.t}{f_{k}r.\frac{\omega }{2}.t}=\frac{\frac{1}{2}mv+v_{0}v-v_0}{\frac{1}{2}.\frac{2}{5}m{r}^2{\omega }^2}\Rightarrow -1\frac{v-v_0}{\frac{2}{5}\omega r}$

But $r\omega =v$.

$\Rightarrow -1=\frac{v-v_0}{\frac{2}{5}\omega r}$

$\Rightarrow -\frac{2}{5}v=v-v_0\Rightarrow v_0=v+\frac{2}{5}v=\frac{7}{5}v\Rightarrow v=\frac{5}{7}{v}_0$

3. By conservation of angular momentum:

if $\overrightarrow{tau}=0,\frac{\overrightarrow{d}L}{dt}=0,L=constant$

The torque acting on the ball for all its positions, about point P (a point on the ground as shown in the figure) is zero. So the angular momentum of the ball about point P will remain constant.

$\Rightarrow m{v}_{0}r=I_{cm}\omega +mvr\Rightarrow m{v}_{0}r=\frac{2}{5}m{r}^2\left(\frac{v}{r}\right)+mvr(\omega =(\frac{v}{r}\left)\right)$

$\Rightarrow v_0=\frac{2}{5}v+v\Rightarrow v=\frac{5}{7}{v}_0$