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

Three solid spheres are made to move on a rough horizontal surface. Sphere $P$ is given a spin and released. Sphere $Q$ is given a forward linear velocity. Sphere $R$ is given linear and rotational motions as shown in the figure. Directions of the friction force on spheres $P, Q, R$ are respectively

$A$ ball is thrown on a lawn in such a way that it initially slides with a speed $v_0$ without rolling. It gradually picks up rotational motion. The speed of the ball at which there will be rolling without slipping is

In a bicycle,the radius of the rear wheel is twice the radius of the front wheel. If $r_F$ and $r_r$ are the radii,and $v_F$ and $v_r$ are the speeds of the topmost points of the wheels respectively,then:

The speed of rolling of a ring of mass $M$ changes from $V$ to $3\ V$. What is the change in its kinetic energy (in $,MV^2$)?

$A$ circular disc of mass $2 \,kg$ and radius $10 \,cm$ rolls without slipping with a speed $2 \,m/s$. The total kinetic energy of the disc is .......... $J$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module