Question

A heavy solid sphere is thrown on a horizontal rough surface with initial velocity $u$ without rolling. What will be its speed when it starts pure rolling? (Assume $u=7\text{m/s}$)

• A. $4.2\text{m/s}$
• B. $5\text{m/s}$
• C. $5.6\text{m/s}$
• D. $3.5\text{m/s}$

Answer 1

The correct option is B $5\text{m/s}$

Given,
Initial linear velocity $=u=7\text{m/s}$
Initial angular velocity $\omega =0$


Initially due to slipping, friction will act opposite to the direction of motion, due to which torque will act. This will produce angular acceleration (in clockwise dirction). The angular speed starts increasing and linear speed starts decreasing.

After some time sphere starts pure rolling. Let its linear velocity be $v$ and angular velocity ${\omega }^{\prime }$.
Then, $v={\omega }^{\prime }R...\left(1\right)$
In that situation, there will be no sliding between sphere and ground.
Hence, $f_k=0,\alpha =0$
${\omega }^{\prime }=\text{constant},v=\text{constant}$

By conservation of angular momentum about point $P$ $({\tau }_{net}=0$)
$L_i=L_f$
$\Rightarrow -muR=-mvR-I{\omega }^{\prime }$
$\Rightarrow muR=mvR+\left(\frac{2}{5}m{R}^2\right)\left(\frac{v}{R}\right)$
$\Rightarrow u=\frac{7}{5}v$
$\Rightarrow v=\frac{5u}{7}=5\text{m/s}$