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

$A$ solid sphere of mass $1 \ kg$ rolls on a table with linear speed $1 \ m/s$. Its total kinetic energy is .......... $J$.

Which of the following statements is not correct?

$A$ wheel of radius $r$ rolls without slipping with a speed $v$ on a horizontal road. When it is at a point $A$ on the road,a small piece of mud separates from the wheel at its highest point $B$ and drops at point $C$ on the road. The distance $AC$ will be

$A$ thin circular ring and a circular disc of equal mass are rolling without sliding. If their linear velocities are equal and the total kinetic energy of the disc is $6 \ J$,then the total kinetic energy of the ring is (in $J$)

$A$ 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 $J$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module