Rotational Dynamics of a meter stick

In summary, the problem involves finding the moment of inertia of a meter stick with a mass of 0.44 kg rotating horizontally about a vertical axis passing through the 30 cm mark. The correct formula to use is I=1/12(M)(L^2) for a long uniform rod. To find the moment of inertia about the 30 cm mark, the parallel axis theorem is used by adding the distance between the new axis and the center of mass to the moment of inertia through the center of mass. The final answer is I=0.0762 kg/m.
  • #1
Lma12684
25
0

Homework Statement


A meter stick of mass .44 kg rotates, in the horizontal plane, about a vertical axis passing thru the 30 cm mark. What is the moment of inertia of the stick? Treat it as a long uniform rod.



Homework Equations


I=1/2M(R^2)

The Attempt at a Solution


=1/2(.44)(.3^2)

Once I calculate this, is the the problem complete? It seems too simple.
 
Physics news on Phys.org
  • #2
Hi Lma12684,

The formula I=(1/2)MR^2 is not the moment of inertia for a meter stick. Different shapes have different moment of inertia formulas (although these formulas all are derived from the same integral definition).

You would probably want to use the formula for moment of inertia of a rod here (this assumes that the width of the meterstick is much less than its length, which is probably reasonable). You can probably find a formula in a table for a rod with a rotational axis through the center of the rod and another formula with the axis through one end.

Once you have that, how do you use that expression to find [itex]I[/itex] for an axis at 30cm?
 
  • #3
Should I assume that 30 cm is the middle of the rod? Thanks.
 
  • #4
No, you can't do that since it's a meter stick. But I would first find the formula that applies when it is rotating about the center of the rod. What do you get for that case?

Once you have that, you can use that formula with the parallel axis theorem to find the moment of inertia for the case of a rotation axis that is not at the center.
 
  • #5
Relavent formula: I=(sum of t)/angular acceleration
 
  • #6
Do you have your textbook available? If so, I think you should be able to find a table that has some shapes and the formulas for their moments of inertia. (There will probably be some for spheres, rods, disks, etc.) You'll probably be using that table quite a bit so it's important to locate it.

If you don't have your book available, you can look here:

http://en.wikipedia.org/wiki/List_of_moments_of_inertia


(The formula you quoted I=(sum of t)/angular acceleration is a general relationship that is the angular analogue of Newton's law.)

Once you have that moment of inertia for a rotation about the 50cm mark, then the parallel axis theorem will allow you to find it at the 30cm mark.
 
  • #7
My mistake, I misunderstood you. The formula I found for a long uniform rod was I=1/12(M)(L^2).

For L, so I consider it as 1 meter? Or do I use .5 m?

I am not sure what you mean by parellel axis theorem.

Thanks!
 
  • #8
In that formula the L is the total length of the rod.

I would think the parallel axis theorem is in your textbook. Once you find the moment of inertia through the center of mass, it allows you to find the moment of inertia about any axis that is parallel to that; all you need is the mass and the distance between the old and new axes. What do you get?
 
  • #9
I found that I=.0366 kg/m.

Now, I used I=I(cm) + m(d^2).
.0366=I + (.44)(.3^2)
I=-.003

I know I did something wrong. Did I use the wrong distance?
 
  • #10
The value you found (0.0366) was the moment of inertia through the center of mass [itex]I_{\rm cm}[/itex], and so goes on the right side of the equation. You are solving for the new I in this equation.

Also, the d is the distance between the two axes. How far is the center of mass and the new axis? It's not 30cm.
 
  • #11
I think I figured it out. I had my I(cm) in the wrong place of the equation. I now found that I=.0762. THanks!
 
  • #12
Sorry, I just saw your post. Let me check it again. Thanks.
 

What is rotational dynamics?

Rotational dynamics is the study of the motion of objects that rotate or spin around a fixed axis. It involves the application of Newton's laws of motion and principles of torque and angular momentum.

How does rotational motion differ from linear motion?

Rotational motion involves movement around a fixed axis, while linear motion occurs in a straight line. This means that rotational motion has a specific direction and involves angular displacement, velocity, and acceleration, while linear motion has a single direction and involves only linear displacement, velocity, and acceleration.

What is the role of torque in rotational dynamics?

Torque is the measure of the force that causes an object to rotate around an axis. It is calculated by multiplying the force applied to an object by the distance from the axis of rotation. In rotational dynamics, torque is used to determine the angular acceleration of an object.

How is angular momentum related to rotational dynamics?

Angular momentum is the measure of an object's tendency to continue rotating around an axis. In rotational dynamics, it is conserved, meaning it remains constant unless acted upon by an external torque. This principle is often used to analyze objects in rotational motion, such as a meter stick.

What are some real-world applications of rotational dynamics?

Rotational dynamics has many practical applications, including in the design of machines and vehicles that involve rotational motion, such as engines and turbines. It is also used in sports and games, such as in calculating the spin of a ball in baseball or the rotation of a figure skater. Additionally, rotational dynamics is important in understanding the stability and motion of celestial bodies, such as planets and stars.

Similar threads

  • Introductory Physics Homework Help
3
Replies
95
Views
3K
  • Introductory Physics Homework Help
Replies
9
Views
5K
Replies
7
Views
236
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
19
Views
3K
  • Introductory Physics Homework Help
Replies
14
Views
2K
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top