Moment of Inertia and Linear/Angular Momentum

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Homework Help Overview

The discussion revolves around a problem involving the moment of inertia and the conservation of linear and angular momentum in a system consisting of a disk and a wad of clay. The scenario describes a thin disk rotating about a vertical axis and a clay projectile impacting it.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore calculations for the moment of inertia of both the clay and the disk, as well as the linear and angular momentum before and after the impact. There is an attempt to identify errors in the calculations, particularly regarding the mass of the clay.

Discussion Status

Some participants have pointed out a mistake in the mass of the clay, which was initially stated as 1.0 kg instead of the correct 1.0 g. This correction has led to further clarification and adjustments in the calculations presented.

Contextual Notes

The problem includes specific constraints regarding the mass and dimensions of the objects involved, which are critical for accurate calculations. The discussion reflects an ongoing effort to reconcile these values with the physics principles at play.

mparsons06
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1. Homework Statement .

A very thin 1.0 kg disk with diameter 80 cm is mounted horizontally to rotate freely about a central vertical axis. On the edge of the disk, sticking out a little, is a small, essentially mass-less tab, or "catcher." A 1.0 g wad of clay is fired at a speed of 10.0 m/s directly at the tab, perpendicular to it, and tangent to the disk. The clay sticks to the tab, which is initially at rest.

A.) What is the moment of inertia of the clay about the axis? (kg·m^2)

B.) What is the moment of inertia of the disk about the axis? (kg·m^2)

C.) What is the moment of inertia of both the clay and the disk about the axis? (kg·m^2)

D.) What is the linear momentum of the clay before impact? (kg·m^2)

E.) What is the angular momentum of the clay just before impact? (kg·m^2)

F.) What is the angular speed of the disk after impact? (rad/s)


2. The attempt at a solution.

A.) What is the moment of inertia of the clay about the axis? (kg•m^2)
I clay = mass clay * radius^2
I clay = (0.001 kg) * (0.4 m)^2
I clay = 0.00016 kg•m^2

B.) What is the moment of inertia of the disk about the axis? (kg•m^2)
I disk = ½ * mass disk * radius^2
I disk = ½ * (1.0 kg) * (0.4 m)^2
I disk = 0.08 kg•m^2

C.) What is the moment of inertia of both the clay and the disk about the axis? (kg•m^2)
I clay + I disk = mass clay * radius^2 + ½ * mass disk * radius^2
I total = radius^2 (mass clay + ½ * mass disk)
I total = (0.4 m)^2 * (0.001 kg + ½ * 1.0 kg)
I total = 0.08 kg•m^2

D.) What is the linear momentum of the clay before impact? (kg•m^2)
p = mass * velocity
p = (1.0 kg) * (10.0 m/s)
p = 10.0 kg•m^2

E.) What is the angular momentum of the clay just before impact? (kg•m^2)

Angular velocity = ω = radians/s
Radian = length of radius
r = 0.4 m
(10 m/s) ÷ (0.4 m/rad) = 25 rad/s = ω

Angular Momentum = I * ω
I clay = mass * radius^2
Angular Momentum = mass clay * radius^2 * 25 rad/s
Initial Angular Momentum clay = (0.001 kg) * (0.4 m)^2 * (25 rad/s) = 0.004 kg•m^2)

F.) What is the angular speed of the disk after impact? (rad/s)

Initial Angular momentum = Final Angular momentum
Initial Angular Momentum = 0.004
Final Angular Momentum = I total * angular velocity final (of clay and disk)

I total = radius^2 *(mass clay + ½ * mass disk)

Final Angular Momentum = radius^2 * (mass clay + ½ * mass disk) * ω
0.004 = radius^2 * (mass clay + ½ * mass disk) * ω
0.004 = (0.4 m)^2 * (0.501 kg) * ω
ω = (0.004) ÷ (0.4 m)^2 * (0.501 kg)
ω = 0.0499001996 rad/s



But D. is wrong. Where did I mess up?
 
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The clay is 1.0 g, not 1.0 kg.
 
Minor mistake. Thanks for pointing it out.
 
kuruman said:
The clay is 1.0 g, not 1.0 kg.

After changing 1.0 g to 0.001 kg * 10 = 0.01 kg.m/s => correct. Thank you ;)
 
Last edited:

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