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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?
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?