A disc is rolling, the velocity of its centre | vedclass

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

Question

(D) For a disc rolling without slipping on a horizontal surface, the velocity of any point on the disc is the vector sum of the velocity of the centre

Vedclass · Accepted Answer

the velocity of highest point is $2 v_{cm}$ and point of contact is zero

Vedclass · Answer

(D) For a disc rolling without slipping on a horizontal surface, the velocity of any point on the disc is the vector sum of the velocity of the centre of mass $(v_{cm})$ and the tangential velocity due to rotation $(v_{rot} = R\omega)$.Since the disc is rolling without slipping, the velocity of the point of contact with the ground is zero, which implies $v_{cm} = R\omega$.At the highest point, the velocity is $v_{top} = v_{cm} + R\omega = v_{cm} + v_{cm} = 2v_{cm}$.At the point of contact, the velocity is $v_{contact} = v_{cm} - R\omega = v_{cm} - v_{cm} = 0$.Therefore, the velocity of the highest point is $2v_{cm}$ and the velocity of the point of contact is zero.

$A$ disc is rolling, the velocity of its centre of mass is $v_{cm}$. Which one will be correct?

Similar Questions

$A$ solid sphere of radius $10 \ cm$ and mass $500 \ g$ is rolling without slipping with a velocity of $20 \ cm/s$. The total kinetic energy of the sphere is: (in $J$)

$A$ solid sphere rolls without slipping on a rough surface and the centre of mass has a constant speed $v_0$. If the mass of the sphere is $m$ and its radius is $R$,then find the angular momentum of the sphere about the point of contact $P$.

The ratio of the total kinetic energy to the rotational kinetic energy for a rolling disc is:

$A$ ring and a disc roll on a horizontal surface without slipping with the same linear velocity. If both have the same mass and the total kinetic energy of the ring is $6 \ J$,then the total kinetic energy of the disc is: (in $/2 \ J$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module