Question

What is the moment of inertia of an object that rolls without slipping down a 2.02 m-high incline starting from rest, and has a final velocity of 5.72 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. MR2

Answer 1

Answer:

$I=0.210\ MR^2$

Explanation:

Given that,

Height of the inclined plane, h = 2.02 m

Final velocity of the object, v = 5.72 m/s

To find,

The moment of inertia of the object.

Solution,

As the object rolls down the inclined plane, its gravitational potential energy gets converted to the kinetic energy along with rotational kinetic energy.

$Mgh=\dfrac{1}{2}Mv^2+\dfrac{1}{2}I\omega^2$

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

$Mgh=\dfrac{1}{2}Mv^2+\dfrac{1}{2}I(\dfrac{v}{R})^2$

On rearranging the above equation and finding the value of I as :

$I=(\dfrac{2gh}{v^2}-1)\ MR^2$

$I=(\dfrac{2\times 9.8\times 2.02}{(5.72)^2}-1)\ MR^2$

$I=0.210\ MR^2$

So, the moment of inertia of an object that rolls without slipping down is $0.210\ MR^2$