MCQ
A solid sphere spinning about a horizontal axis with an angular velocity $\omega$ is placed on a horizontal surface. Subsequently it rolls without slipping with an angular velocity of
- A$2 \omega / 5$
- B$7 \omega / 5$
- ✓$2 \omega / 7$
- D$\omega$
✓
Answer
Correct option: C.
$2 \omega / 7$
c
(c)
(c)
Initially slipping occurs at point of contact and friction acts in forward direction.
Friction produces acceleration of centre of mass.
Acceleration of centre of mass caused by friction is
$a_{ CM }=\frac{F}{m}=\frac{\mu_k m g}{m}=\mu_k g$
Angular deacceleration caused by friction is
$\alpha=\frac{\tau}{I}=\frac{f R}{I}=\frac{-\mu_k m g R}{\frac{2}{\tilde{\partial}} m R^2}=-\frac{5 \mu_k g}{2 R}$
After time $t$, velocity of centre of mass is
$v_{ CM }=u_{ CM }+a_{ CM } t=0+\mu_k g t=\mu_k g t$
and angular velocity is
$\omega=\omega_0+\alpha t=\omega_{0}-\frac{5 \mu_k g}{2 R}, t$
Pure rolling begins
when $\omega-\frac{v_{ CM }-\mu_k g t}{R} \frac{R}{R-\mu_k}$
So, $\quad \frac{\mu_k g t}{R}=\omega_0-\frac{5}{2} \frac{\mu_k g t}{R}$
$\Rightarrow \quad \frac{7}{2} \frac{\mu_k g t}{R}=\omega \Rightarrow t=\frac{2 R \omega_0}{7 \mu_k g}$
So, angular speed when pure rolling occurs is
$\omega_f=\omega+\alpha t=\omega-\frac{5 \mu_k g}{2 R} \cdot \frac{2 R \omega}{7 \mu_k g}$
$=\omega\left(1-\frac{5}{7}\right)=\frac{2}{7} \omega$
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