A body is rolling without slipping on a horiz | vedclass

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

Question

(D) For rolling without slipping, the velocity is $v = r \omega$. Total kinetic energy $E = K_{rot} + K_{trans} = \frac{1}{2} I \omega^2 + \frac{1}{2}

Vedclass · Accepted Answer

Thin circular ring

Vedclass · Answer

(D) For rolling without slipping, the velocity is $v = r \omega$. Total kinetic energy $E = K_{rot} + K_{trans} = \frac{1}{2} I \omega^2 + \frac{1}{2} m v^2$. Substituting $v = r \omega$, we get $E = \frac{1}{2} I \omega^2 + \frac{1}{2} m r^2 \omega^2 = \frac{1}{2} (I + m r^2) \omega^2$. Rotational kinetic energy is $K_{rot} = \frac{1}{2} I \omega^2$. Given that $K_{rot} = 50 \% 	ext{ of } E$, so $K_{rot} = \frac{1}{2} E$. Substituting the expressions: $\frac{1}{2} I \omega^2 = \frac{1}{2} [\frac{1}{2} (I + m r^2) \omega^2]$. $I = \frac{1}{2} I + \frac{1}{2} m r^2$. $\frac{1}{2} I = \frac{1}{2} m r^2$, which implies $I = m r^2$. The moment of inertia $I = m r^2$ corresponds to a thin circular ring.

$A$ body is rolling without slipping on a horizontal plane. If the rotational kinetic energy of the body is $50 \%$ of its total kinetic energy, then the body is

Similar Questions

$A$ sphere of mass $M$ and radius $r$ slips on a rough horizontal plane. At some instant,it has translational velocity $V_0$ and rotational velocity about the centre $\frac{V_0}{2r}$. The translational velocity when the sphere starts pure rolling motion is

$A$ disc of mass $M$ and radius $R$ is rolling on a horizontal surface with an angular speed $\omega$. The angular momentum of the disc about the origin $O$ is:

In the figure,$E$ and $v_{cm}$ represent the total energy and speed of the centre of mass of an object of mass $1 \ kg$ in pure rolling. The object is

In the case of pure rolling,what will be the velocity of point $A$ of the ring of radius $R$?

Two spheres are rolling with the same velocity (for their $C.M.$). Their ratio of kinetic energy is $2:1$ and the ratio of their radii is $2:1$. Their mass ratio will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module