A thin uniform circular disc rolls with a con | vedclass

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

Question

(A) For a rolling disc of mass $M$ and radius $R$ moving with velocity $v$ and angular velocity $\omega = v/R$,the translational kinetic energy is $K_

Vedclass · Accepted Answer

$3$ times its rotational kinetic energy

Vedclass · Answer

(A) For a rolling disc of mass $M$ and radius $R$ moving with velocity $v$ and angular velocity $\omega = v/R$,the translational kinetic energy is $K_t = \frac{1}{2} Mv^2$. The rotational kinetic energy about the center of mass is $K_r = \frac{1}{2} I \omega^2$. For a disc,the moment of inertia $I = \frac{1}{2} MR^2$. Substituting $I$ and $\omega = v/R$,we get $K_r = \frac{1}{2} (\frac{1}{2} MR^2) (v/R)^2 = \frac{1}{4} Mv^2$. The total kinetic energy $K = K_t + K_r = \frac{1}{2} Mv^2 + \frac{1}{4} Mv^2 = \frac{3}{4} Mv^2$. Comparing $K$ with $K_r$: $K = \frac{3/4 Mv^2}{1/4 Mv^2} K_r = 3 K_r$. Thus,the total kinetic energy is $3$ times its rotational kinetic energy.

$A$ thin uniform circular disc rolls with a constant velocity without slipping on a horizontal surface. Its total kinetic energy is

Similar Questions

$A$ sphere is rolling without slipping on a fixed horizontal plane surface. $A$ is the point of contact,$B$ is the centre of the sphere,and $C$ is the topmost point. Then:

An object of mass $10 \ kg$ and radius $0.5 \ m$ is rolling without slipping with a velocity of $2 \ m/s$. Its total kinetic energy is $32.8 \ J$. The radius of gyration of the object is .......... $m$.

$A$ force $F$ is applied at the centre of a disc of mass $M$. The minimum value of the coefficient of friction of the surface for pure rolling is:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module