A solid uniform sphere resting on a rough hor | vedclass

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

Question

(C) Let $v$ be the final velocity of the centre of the sphere and $\omega$ be the final angular velocity when it starts rolling without slipping.Since

Vedclass · Accepted Answer

$\frac{5}{7} v_{0}$

Vedclass · Answer

(C) Let $v$ be the final velocity of the centre of the sphere and $\omega$ be the final angular velocity when it starts rolling without slipping.Since the friction force acts at the point of contact,the net torque about the point of contact is zero.Therefore,the angular momentum about the point of contact remains conserved.Initial angular momentum about the point of contact: $L_i = m v_0 r$Final angular momentum about the point of contact: $L_f = mvr + I_{cm}\omega$For a solid sphere,the moment of inertia about its centre is $I_{cm} = \frac{2}{5} mr^2$.Since it rolls without slipping,the condition is $v = r\omega$,or $\omega = \frac{v}{r}$.Equating initial and final angular momentum:$mv_0 r = mvr + (\frac{2}{5} mr^2)(\frac{v}{r})$$mv_0 r = mvr + \frac{2}{5} mvr$$v_0 = v + \frac{2}{5} v$$v_0 = \frac{7}{5} v$$v = \frac{5}{7} v_0$

$A$ 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

Similar Questions

$A$ particle on the outer surface of a wheel,which is initially at rest,is in contact with the ground at point $P$. Find the displacement of this particle when the wheel completes half a rotation in the forward direction. (Radius of the wheel $= 5 \ m$)

Explain why friction is necessary to make the disc in the figure roll in the direction indicated.
$(a)$ Give the direction of frictional force at $B$,and the sense of frictional torque,before perfect rolling begins.
$(b)$ What is the force of friction after perfect rolling begins?

$A$ circular ring of mass $10 \text{ kg}$ rolls along a horizontal floor. The center of mass of the ring has a speed of $1.5 \text{ m/s}$. The work required to stop the ring is: (in $\text{ J}$)

$A$ uniform solid sphere of radius $R$ and radius of gyration $K$ about an axis passing through the centre of mass,is rolling without slipping. Then the fraction of total energy associated with its rotation will be

$A$ uniform solid sphere of mass $m$ and radius $R$ is being pulled on a horizontal surface with a force $F$ parallel to the surface,applied at its topmost point. If the acceleration of the center of mass of the sphere is $a$ and it is rolling without slipping on the surface,the value of $F$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module