Understanding Rigid Body Rotation: Solving Problem 4A and B from a Physics Exam

In summary, the moment of inertia for a cylinder (representing a person) includes (1/2)MR^2, with the arms hanging down. When the arms are outstretched, the moment of inertia equation includes m(1l/2 + R)^2, with the axis of rotation at the center of the person. The solution provided in the link distinguishes between M and m, and the moment of inertia of the arms is only included in the hanging down case.
  • #1
Kenneth Dirk
5
0
I am trying to solve problem number 4 part A and B from (http://people.physics.tamu.edu/kamon/teaching/phys218/exam/2003C/2003C_Exam3_Solution.pdf) but I am confused about certain aspects of it.

In part A, I understand that since we are considering the person as a cylinder, the equation for moment of inertia will include (1/2)MR^2. What I can't seem to understand is where did 2MR^2 come from and why we didn't include the moment of inertia of the arm(which is a rod).

In part B, I understand the rest of the moment of inertia equation but I am confused where did m(1l/2 + R)^2 come from?
 
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  • #2
Hello Ken, :welcome:

Please use the template and the superscript/subscript buttons. See guidelines
Apparently you have a question on moment of inertia. So isolate that problem and state it. Show the relevant equations.

And note that the solution does make a difference between M and m -- as you do not.
 
  • #3
Kenneth Dirk said:
What I can't seem to understand is where did 2MR^2 come from and why we didn't include the moment of inertia of the arm(which is a rod).
The 2mR^2 is the moment of inertia of the arms -- when they are hanging down, not outstretched. (Note: m, not M.)

Kenneth Dirk said:
In part B, I understand the rest of the moment of inertia equation but I am confused where did m(1l/2 + R)^2 come from?
Here the arms are outstretched. Note that the axis of rotation is the center of the person, not the shoulder.
 
  • #4
Doc Al said:
The 2mR^2 is the moment of inertia of the arms -- when they are hanging down, not outstretched. (Note: m, not M.)Here the arms are outstretched. Note that the axis of rotation is the center of the person, not the shoulder.
Giving it away, eh ? not PF ! Well, let's say Ken has beginner's credit :wink:
 
  • #5
BvU said:
Giving it away, eh ? not PF !
:wink:

BvU said:
Well, let's say Ken has beginners credit
First one's free! :smile:
 
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Likes BvU

1. What is a rigid body rotation problem?

A rigid body rotation problem is a physics problem that involves the rotation of a solid object around an axis. The object is considered to be rigid if its shape and size do not change during the rotation.

2. What are the key concepts involved in solving a rigid body rotation problem?

The key concepts involved in solving a rigid body rotation problem include torque, angular momentum, and moment of inertia. Torque is the force that causes rotational motion, angular momentum is the measure of an object's rotation, and moment of inertia is the object's resistance to changes in its rotation.

3. How do you find the moment of inertia in a rigid body rotation problem?

The moment of inertia can be calculated by multiplying the mass of the object by the square of its distance from the axis of rotation. The moment of inertia is also affected by the distribution of mass in the object.

4. What are the different types of rigid body rotations?

There are two types of rigid body rotations: rotational motion about a fixed axis and rotational motion about a moving axis. In rotational motion about a fixed axis, the axis of rotation remains constant, while in rotational motion about a moving axis, the axis of rotation changes as the object rotates.

5. How do you solve a rigid body rotation problem?

To solve a rigid body rotation problem, you can use the equations of rotational motion, which relate the angular acceleration, angular velocity, and angular displacement of the object. You can also use the conservation of angular momentum and the parallel axis theorem to simplify the problem and find the solution.

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