Question

You have a pumpkin of mass M and radius R. The pumpkin has the shape of a sphere, but it is not uniform inside so you do not know its moment of inertia. In order to determine the moment of inertia, you decide to roll the pumpkin down an incline that makes an angle θ with the horizontal. The pumpkin starts from rest and rolls without slipping. When it has descended a vertical height H, it has acquired a speed of v = √(5gH/4). Le, Tim - le228@m

Answer 1

The moment of inertia (J) is
`\(J = (3)/(5)MR^2\)`

How to determine the moment of inertia?

When an object descends from a height \(h\), it loses potential energy,
`\(Mgh\)`, and gains kinetic energy,
`\((1)/(2)Mv^2\)`, where
`\(M\)` is the mass of the object and
`\(v\)` is its velocity.

Additionally, it acquires rotational kinetic energy,
`\((1)/(2)Jw^2\)`, where
`\(J\)` is the moment of inertia and
`\(w\)` is the angular speed (
`\(w = (v)/(R)\), with \(R\)` being the radius of rotation).

By equating the energies:
`\(Mgh = (1)/(2)Mv^2 + (1)/(2)J\left((v)/(R)\right)^2\)`, and further simplifying:

`\(2gh/v^2 = 1 + (J)/(MR^2)\)`

We know
`\(v = \sqrt{(5gh)/(4)}\)`. Thus,
`\((J)/(MR^2) = 2gh/v^2 - 1\)`.

Solving for
`\((J)/(MR^2)\)`:
`\((J)/(MR^2) = (8gh)/(5gh) - 1\)`, therefore
`\((J)/(MR^2) = (3)/(5)\)`.

Consequently,
`\(J = (3)/(5)MR^2\)`.

Answer 2

`I = 1.6MR^2`

Step-by-step explanation:

When the pumpkin rolling down an incline of vertical height H, its potential energy is converted to rotational energy and kinetic energy

`MgH = (Mv^2)/(2) + (I\omega^2)/(2)`

where
`\omega = (v)/(R)` is the angular speed

`2MgH - Mv^2 = (Iv^2)/(R^2)`

`2MgHR^2 - Mv^2R^2 = Iv^2`

`I = 2MgH(R^2)/(v^2) - MR^2`

we can substitute
`v = \sqrt{(5gH)/(4)}`

`v^2 = (5gH)/(4)`

`I = 8MgH(R^2)/(5gH) = 1.6MR^2`