If the rotational kinetic energy is 50% of th | vedclass

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

Question

(A) The total kinetic energy $(K_{total})$ is the sum of translational kinetic energy $(K_t)$ and rotational kinetic energy $(K_r)$.Given that $K_r =

Vedclass · Accepted Answer

Ring

Vedclass · Answer

(A) The total kinetic energy $(K_{total})$ is the sum of translational kinetic energy $(K_t)$ and rotational kinetic energy $(K_r)$.Given that $K_r = 50\% 	ext{ of } K_{total}$,it implies $K_r = K_t$.We know $K_r = \frac{1}{2} I \omega^2$ and $K_t = \frac{1}{2} mv^2$.Equating them: $\frac{1}{2} I \omega^2 = \frac{1}{2} mv^2$.Since $v = \omega r$,we have $\frac{1}{2} I \omega^2 = \frac{1}{2} m(\omega r)^2$.Simplifying,$I \omega^2 = m \omega^2 r^2$,which gives $I = mr^2$.The moment of inertia $I = mr^2$ corresponds to a ring.

If the rotational kinetic energy is $50\%$ of the total kinetic energy,then the body is a .......... .

Similar Questions

$A$ ring of mass $m$ and radius $r$ rotates about an axis passing through its centre and perpendicular to its plane with angular velocity $\omega$. Its kinetic energy is

If the rotational kinetic energy of a body is increased by $300\ \%$,then the percentage increase in its angular momentum will be .......... $\%$

If $L, M$,and $P$ are the angular momentum,mass,and linear momentum of a particle respectively,which of the following represents the kinetic energy of the particle when the particle rotates in a circle of radius $R$?

$A$ body is rotating about its own axis. Its rotational kinetic energy is $x$ and its angular momentum is $y$. Hence,its moment of inertia about its own axis is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module