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

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PhantomWisdom PhantomWisdom · Answer 1 · 2019-11-10T09:01:17+01:00

Answer:

[tex]\dfrac{T}{K}=1[/tex]

Explanation:

The translational kinetic energy of the hoop is given by :

[tex]K=\dfrac{1}{2}Mv^2[/tex]..................(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 :

[tex]T=\dfrac{1}{2}I\omega^2[/tex]

Since, [tex]I=MR^2[/tex]

[tex]\omega=\dfrac{v}{R}[/tex]

[tex]T=\dfrac{1}{2}\times MR^2\times (\dfrac{v}{R})^2[/tex]..............(2)

From equation (1) and (2) :

[tex]\dfrac{T}{K}=1[/tex]

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