Question

g hoop rolls without slipping on a horizontal surface. The hoop has a mass m and radius r. What is the ratio of the hoop's translational kinetic energy to the hoop's rotational kinetic energy, KT /KR

Answer 1

Answer:

$\frac{K.E_t}{K.E_r} = 1$

Explanation:

The translational kinetic energy of the hoop is given as:

$K.E_t = \frac{1}{2} mv^2$      ---------------------- equation (1)

where,

$K.E_t$ = translational kinetic energy

m = mass of hoop

v = linear speed of hoop

The rotational kinetic energy of the hoop is given as:

$K.E_r = \frac{1}{2} I\omega^2$

where,

$K.E_r$ = rotational kinetic energy of the hoop

I = Moment of Inertia of the hoop = mr²

r = radius of the hoop

ω = angular speed of hoop = $\frac{v}{r}$

Therefore,

$K.E_r = \frac{1}{2} (mr^2)(\frac{v}{r} )^2\\\\K.E_r = \frac{1}{2} mv^2$------------------- equation (2)

dividing equation (1) and equation (2), we get:

$\frac{K.E_t}{K.E_r} = \frac{\frac{1}{2}mv^2 }{\frac{1}{2}mv^2 }\\\\\frac{K.E_t}{K.E_r} = 1$