A solid sphere spinning about a horizontal ax | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Initially,the sphere is placed on the surface with angular velocity $\omega$ and linear velocity $v_{CM} = 0$. Friction acts at the point of conta

Vedclass · Accepted Answer

$2 \omega / 7$

Vedclass · Answer

(C) Initially,the sphere is placed on the surface with angular velocity $\omega$ and linear velocity $v_{CM} = 0$. Friction acts at the point of contact to oppose the slipping.The force of friction $f = \mu_k mg$ acts in the forward direction,causing linear acceleration $a_{CM} = \frac{f}{m} = \mu_k g$.The torque due to friction about the center of mass is $	au = -fR = -\mu_k mgR$. The angular deceleration is $\alpha = \frac{	au}{I} = \frac{-\mu_k mgR}{\frac{2}{5}mR^2} = -\frac{5\mu_k g}{2R}$.After time $t$,the linear velocity is $v_{CM} = a_{CM}t = \mu_k gt$ and the angular velocity is $\omega_f = \omega + \alpha t = \omega - \frac{5\mu_k g}{2R}t$.Pure rolling begins when $v_{CM} = \omega_f R$. Substituting the expressions:$\mu_k gt = \left(\omega - \frac{5\mu_k g}{2R}tight)R$$\mu_k gt = \omega R - \frac{5}{2}\mu_k gt$$\frac{7}{2}\mu_k gt = \omega R \implies \mu_k gt = \frac{2}{7}\omega R$.Substituting this back into the expression for $\omega_f$:$\omega_f = \frac{v_{CM}}{R} = \frac{\mu_k gt}{R} = \frac{1}{R} \left(\frac{2}{7}\omega Right) = \frac{2}{7}\omega$.

$A$ 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

Similar Questions

$A$ solid uniform sphere resting on a rough horizontal plane is given a horizontal impulse directed through its centre so that it starts sliding with an initial velocity $v_{0}$. When it finally starts rolling without slipping,the speed of its centre is

$A$ thin hollow cylinder,open at both ends,slides without rotating and then rolls without slipping with the same speed. The ratio of kinetic energies in the two cases will be:

$A$ solid sphere of mass $1\,kg$ rolls without slipping on a plane surface. Its kinetic energy is $7 \times 10^{-3}\,J$. The speed of the centre of mass of the sphere is $.........\,cm\,s^{-1}$.

$A$ sphere of mass $50\,g$ and diameter $20\,cm$ rolls without slipping with a velocity of $5\,cm/s$. Its total kinetic energy is

When a body is rolling without slipping on a rough horizontal surface,the work done by friction is ........

Vedclass Test Series

Exam Paper Generator

Online Exam Module