A wheel is rolling on a plane surface. The sp | vedclass

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

Question

(A) For a wheel rolling on a plane surface,the velocity of the highest point $B$ is $V_B = 2v$,where $v$ is the velocity of the center of mass.Given $

Vedclass · Accepted Answer

$4 \sqrt{2} \ m/s$

Vedclass · Answer

(A) For a wheel rolling on a plane surface,the velocity of the highest point $B$ is $V_B = 2v$,where $v$ is the velocity of the center of mass.Given $V_B = 8 \ m/s$,we have $2v = 8 \ m/s$,which implies $v = 4 \ m/s$.The point $A$ at the bottom is the instantaneous center of rotation.The velocity of any point $P$ on the rim is given by $V_P = \omega r_P$,where $r_P$ is the distance from the instantaneous center $A$.For a point $P$ at the same level as the center $C$,the distance $AP = \sqrt{R^2 + R^2} = R\sqrt{2}$,where $R$ is the radius of the wheel.Since $v = \omega R$,we have $\omega = v/R$.Thus,$V_P = (v/R) 	imes (R\sqrt{2}) = v\sqrt{2}$.Substituting $v = 4 \ m/s$,we get $V_P = 4\sqrt{2} \ m/s$.

$A$ wheel is rolling on a plane surface. The speed of a particle on the highest point of the rim is $8 \ m/s$. The speed of the particle on the rim of the wheel at the same level as the centre of the wheel will be

Similar Questions

Find the displacement of the point of contact with the ground when a ring of radius $R$ completes a half-rotation.

$A$ solid sphere is rolling with a linear velocity $v$. Its total kinetic energy is:

Why are the contact points of the surface of a circular body stationary during pure rolling motion?

Can the relation $v = r\omega$ be used for an object undergoing rolling motion with slipping? Why?

Vedclass Test Series

Exam Paper Generator

Online Exam Module