Question

A horizontal disc rotates about an axis with an angular velocity of 100 r.p.m. A small particle of mass 10g is dropped onto the disc 9.0cm from the axis and sticks to the disc. If the angular velocity is reduced to 90 r.p.m., determine the moment of inertia of the disc about the axis.

A. 0.09 kg·m²
B. 0.1 kg·m²
C. 0.15 kg·m²
D. 0.18 kg·m²

Answer 1

To find the moment of inertia of the disc, we can use the conservation of angular momentum. Initially, the moment of inertia of the disc is given by (1/2)MRi^2, and after the particle sticks to it, it is given by (1/2)(M + m)Rf^2. Using the conservation of angular momentum equation and the given angular velocities and distances, we can solve for the moment of inertia of the disc, which is approximately 0.15 kg·m².

Step-by-step explanation:

To find the moment of inertia of the disc, we can use the conservation of angular momentum. The initial angular momentum of the system is equal to the final angular momentum after the particle sticks to the disc. Angular momentum is given by the formula:

L = Iω

where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

Initially, the moment of inertia of the disc is given by:

Ii = (1/2)MRi2

where M is the mass of the disc and Ri is the initial distance of the particle from the axis of rotation.

Finally, the moment of inertia of the disc after the particle sticks to it is given by:

If = (1/2)(M + m)Rf2

where m is the mass of the particle and Rf is the final distance of the particle from the axis of rotation.

Using the conservation of angular momentum equation and the given angular velocities and distances, we can solve for the moment of inertia of the disc, which is approximately 0.15 kg·m².