Question

What is the moment of inertia of an object that rolls without slipping down a 3.5-m- high incline starting from rest, and has a final velocity of 7.3 m/s? Express the moment of inertia as a multiple of MR2, where M is the mass of the object and R is its radius. (Hint: Use total conservation of mechanical energy)

Answer 1

Answer:

I = 0.287 MR²

Explanation:

given,

height of the object = 3.5 m

initial velocity = 0 m/s

final velocity  = 7.3 m/s

moment of inertia = ?

Using total conservation of mechanical energy

change in potential energy will be equal to change in KE (rotational) and KE(transnational)

PE = KE(transnational) + KE (rotational)

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2$

v = r ω

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\dfrac{Iv^2}{r^2}$

$I = \dfrac{m(2gh - v^2)r^2}{v^2}$

$I = \dfrac{mr^2(2\times 9.8 \times 3.5 - 7.3^2)}{7.3^2}$

$I =mr^2(0.287)$

I = 0.287 MR²