A disc is rolling (without slipping) on a horizontal surface. C is its centre and Q and P are two points equidistant from C. Let vP, vq and vc be the magnitude of velocities of points P,Q and C respectively, then

Question

A disc is rolling (without slipping) on a horizontal surface. C is its centre and Q and P are two points equidistant from C. Let vP, vq and vc be the magnitude of velocities of points P,Q and C respectively, then (A) vQ > vC > vP (B) vQ < vC < vP (C) vQ=vP,vC=(1/2)vP (D) vQ < vC > vP

Tardigrade Admin · Accepted Answer

In case of pure rolling bottom most point is the instantaneous centre of zero velocity. Velocity of any point on the disc, v=r ω, where r is the distance of point from O rQ > rC > rP ∴ vQ > vC > vP

Q. A disc is rolling (without slipping) on a horizontal surface. C is its centre and Q and P are two points equidistant from C. Let $v_{P}$ , $v_{q}$ and $v_{c}$ be the magnitude of velocities of points P,Q and C respectively, then

$v_{Q} > v_{C} > v_{P}$

$v_{Q} < v_{C} < v_{P}$

$v_{Q} = v_{P}, v_{C} = \frac{1}{2} v_{P}$

$v_{Q} < v_{C} > v_{P}$

In case of pure rolling bottom most point is the instantaneous centre of zero velocity.
Velocity of any point on the disc, $v = r ω$ , where $r$ is the distance of point from $O$
$r_{Q} > r_{C} > r_{P}$
$∴ v_{Q} > v_{C} > v_{P}$

Q. A disc is rolling (without slipping) on a horizontal surface. C is its centre and Q and P are two points equidistant from C. Let vP​, vq​ and vc​ be the magnitude of velocities of points P,Q and C respectively, then

vQ​>vC​>vP​

vQ​<vC​<vP​

vQ​=vP​,vC​=21​vP​

vQ​<vC​>vP​