Solution Found!
a. A disk of mass M and radius R has a hole of radius r
Chapter 12, Problem 56P(choose chapter or problem)
a. A disk of mass \(M\) and radius \(R\) has a hole of radius \(r\) centered on the axis. Calculate the moment of inertia of the disk.
b. Confirm that your answer agrees with Table 12.2 when \(r=0\) and when \(r=R\).
c. A \(4.0-cm\)-diameter disk with a \(3.0-cm\)-diameter hole rolls down a \(50-cm\)-long, \(20^{\circ}\) ramp. What is its speed at the bottom? What percent is this of the speed of a particle sliding down a frictionless ramp?
Equation Transcription:
20o
Text Transcription:
M
R
r
r = 0
r = R
4.0-cm
3.0-cm
50-cm
20 degree
Questions & Answers
QUESTION:
a. A disk of mass \(M\) and radius \(R\) has a hole of radius \(r\) centered on the axis. Calculate the moment of inertia of the disk.
b. Confirm that your answer agrees with Table 12.2 when \(r=0\) and when \(r=R\).
c. A \(4.0-cm\)-diameter disk with a \(3.0-cm\)-diameter hole rolls down a \(50-cm\)-long, \(20^{\circ}\) ramp. What is its speed at the bottom? What percent is this of the speed of a particle sliding down a frictionless ramp?
Equation Transcription:
20o
Text Transcription:
M
R
r
r = 0
r = R
4.0-cm
3.0-cm
50-cm
20 degree
ANSWER:
Step 1 of 5
a.)
We have to calculate the moment of inertia of the given disk.
Let us divide the disk between radii \(r_{1}\) and \(r_{2}\) into narrow rings of mass \(d m\) as shown in the below figure.
The area of a ring of radius \(r\) is \(d A=2 \pi r d r\) and the mass \(d m\) in the ring is the same fraction of the total mass \(M\) as \(d A\) is of the total area \(A\).