Question

A hoop is rolling (without slipping) on a horizontal surface so it has two types of kinetic energy: translational kinetic energy and rotational kinetic energy. The entire mass M of the hoop is concentrated at its rim, so its moment of inertia is I = MR2, where R is the radius. What is the ratio of the translational kinetic energy to the rotational kinetic energy?a) 2, b) 1/4, c) 4, d) 1/2 e) 1

Answer 1

Answer:

$\dfrac{T}{K}=1$

Explanation:

The translational kinetic energy of the hoop is given by :

$K=\dfrac{1}{2}Mv^2$..................(1)

M is the mass of the hoop

v is the velocity of the hoop

The rotational kinetic energy of the hoop is given by :

$T=\dfrac{1}{2}I\omega^2$

Since, $I=MR^2$

$\omega=\dfrac{v}{R}$

$T=\dfrac{1}{2}\times MR^2\times (\dfrac{v}{R})^2$..............(2)

From equation (1) and (2) :

$\dfrac{T}{K}=1$

Therefore, the ratio of the translational kinetic energy to the rotational kinetic energy is 1.