3 Problems and solutions about Moment of inertia equation

1. A solid cylinder has a radius of 8 cm and a mass of 2 kg. Meanwhile, a solid ball has a radius of 5 cm and a mass of 4 kg. If the two objects rotate with an axis through their center, determine the ratio of the moment of inertia of the cylinder and the ball.

__Known:__

Solid cylinder radius (r) = 8 cm = 0.08 m

Solid cylinder mass (m) = 2 kg

Solid ball radius (r) = 5 cm = 0.05 m

The mass of the solid ball (m) = 4 kg

__Wanted:__ Comparison of the moment of inertia of a cylinder and a ball

__Solution____:__

The formula for the moment of inertia of a solid cylinder:

I = ½ M R^{2} = ½ (2)(0,08)^{2 }= 0,0064

The formula for the moment of inertia of a solid ball:

I = 2/5 M R^{2} = 1/5 (4)(0,05)^{2} = (0,8)(0,0025) = 0,002

Comparison of the moment of inertia of the cylinder and the ball:

0,0064 : 0,002

3,2 : 1

2. The tin in which the biscuits were placed was used as a toy. The mass of the can is 200 grams and the radius is 15 cm. The can is rolled on a horizontal floor. If the lid and bottom of the can are neglected, determine the moment of inertia of the can.

__Known:__

The can resembles a hollow cylinder.

Mass (m) = 200 grams = 0.2 kg

Radius (r) = 15 cm = 0.15 m

__Wanted:__

__Solution:__

The formula for the moment of inertia of a hollow cylinder through the axis:

I = M R^{2} = (0,2)(0,15)^{2} = (0,2)(0,0225) = 0,0045 kg m^{2 }

3. A disc which is free to rotate about a vertical axis is capable of rotating at a speed of 80 revolutions per minute. If a small object with a mass of 4 x 10-2 kg is attached to a disc 5 cm from the axis, it turns out that the rotation is 60 revolutions per minute, then determine the moment of inertia of the disc.

__Known:__

Initial angular velocity (ω_{o}) = 80 put / 60 s = 4/3 put/s

Initial moment of inertia (I_{o}) = I

Final angular velocity (ω_{t}) = 60 put / 60 s = 1 put/s

Final moment of inertia (I_{t}) = I + m r^{2 }= I + (0.04)(0.05)^{2} = I + (0.04)(0.0025) = I + 0.0425

__Wanted:__ moment of inertia of the Disc

__Solution:__

The law of conservation of angular momentum

I ω_{o }= I ω_{t}

I (4/3) = I + (0,0425 (1))

4 I / 3 = I + 0,0425

4 I = 3 I + 0,1275

4 I – 3 I = 0,1275

I = 0,1275 kg m^{2}