Question

1. A potter's wheel is rotating around a vertical axis through its center at a frequency of 1.8rev/s . The wheel can be considered a uniform disk of mass 5.0kg and diameter 0.30m . The potter then throws a 3.2-kg chunk of clay, approximately shaped as a flat disk of radius 8.0cm , onto the center of the rotating wheel.

What is the frequency of the wheel after the clay sticks to it? Ignore friction.

Express your answer using two significant figures.

f =		rev/s

Answer 1

Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. Ifthis helps then kindly rate 5-stars.

1

A potter's wheel is rotating around a vertical axis through its center at a frequency of 1.8 rev/s. The wheel can be considered a uniform disk of mass 5.0 kg and diameter 0.40 m. The potter then throws a 3.1 kg chunk of clay that is shaped like a flat disk of radius 8.0 cm, onto the center of the rotating wheel. What is the frequency of the wheel after the clay sticks to it?

Answer

Angular momentum is conserved,

Angular momentum is (inertia * angular velocity)

Inertia for a disk = .5*mass*radius*radius = 0.5 * 5.6 * .2 * .2

The initial angular momentum is then: 0.5 * 5.6 *.2 * .2 * 1.8 (note that is in rev/s)

The inertia for the clay is: 0.5 * 3.1 * .04 * .04

The final angular momentum of disk and clay is then: 0.5 * ((3.1 * .04 * .04) + (5.6 * .2 * .2)) * f where f is the answer you are looking for.

Initial angular momentum = final angular momentum:
0.5 * 5.6 *.2 * .2 * 1.8 = 0.5 * ((3.1 * .04 * .04) + (5.6 * .2 * .2)) * f or

f = (5.6 *.2 * .2 * 1.8)/((3.1 * .04 * .04) + (5.6 * .2 * .2)) in rev/sec

2 A merry-go-round has a mass of 1560 kg and a radius of 7.35 m. How much net work is required to accelerate it from rest to a rotation rate of 1.00 revolution per 7.70 ? Assume it is a solid cylinder.

I am assuming that the meurry-go-round has a rotation rate of 1 revolution per 7.70s.

??=??/?t=(1 revolution)(2? radians/revolution) / (7.70s)= 0.82rad/s

Moment of inertia of a solid cylinder is I=1/2mr²

I=(1/2)(1560kg)(7.35m)²=42137.6kgm²

KE_rotation=1/2I?²

KE_rotation=1/2(42137.6kgm²)(0.82rad/s)²=14166.7 J