Find the rotational inertia of a sphere suspended from the ceiling

In summary, the rotational inertia of a solid uniform sphere of radius R and mass M about a diameter is given by (2/5)MR2. When suspended from the ceiling by a light string of length 2R attached to the surface, the rotational inertia about the point of attachment is 47/5mR^2. The parallel axis theorem is used to solve this problem. For three identical objects fastened to a massless rod of length L, the rotational inertia about one end of the rod is 5/4mL^2. The parallel axis theorem can also be applied here, or one can use the formula for calculating the rotational inertia of a point mass M at a distance D from an axis.
  • #1
grouchy
73
0
A solid uniform sphere of radius R and mass M has a rotational inertia about a diameter that is given by (2/5)MR2. A light string of length 2R is attached to the surface and used to suspend the sphere from the ceiling. Its rotational inertia about the point of attachment at the ceiling is
- 47/5mR^2 (not sure how to find it)


Three identical objects, each of mass M, are fastened to a massless rod of length L as shown. The rotational inertia about one end of the rod of this array is
-5/4mL^2
- again no clue how to get this
 
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  • #2
Consider the parallel axis theorem.
 
  • #3
I looked at that stuff but I don't see how to use to solve these problems.

For the second one I kinda guessed and got it but I don't think its the right way.

I used 1/12ml^2 for the center mass and 1/3ml^2 for the far right mass and added it together. then since there are 3 masses I multiplied by 3 and for 5/4mL^2 but I don't think that's how i"m supposed to find it.
 
  • #4
grouchy said:
I looked at that stuff but I don't see how to use to solve these problems.
Look at it again. For the first problem, you have the rotational inertia about the center of mass but you need it about the point of attachment. Sounds like a perfect opportunity to use the parallel axis theorem!

For the second problem, please attach a diagram.
 
  • #6
Another perfect opportunity to apply the parallel axis theorem. Alternatively: What's the rotational inertia of a point mass M at a distance D from an axis?
 
  • #7
well for the first one I tried

2/5mr^2 + mL^2
2/5mr^2 + 2mL^2 and got 12/5mr^2 which is wrong...honestly I got no clue how this parallel axis thereom stuff works, if u got any tips I'd appreciate it.
 
  • #8
grouchy said:
well for the first one I tried

2/5mr^2 + mL^2

OK, what did you use for "L"? That should be the distance between the center of mass to the point of attachment (in terms of R, of course). (Draw yourself a diagram, unless one is provided--in which case, attach it.)
 
  • #9
I used 2R for L since that's how long it says the string is, and no diagram with the problem :(
 
  • #10
grouchy said:
I used 2R for L since that's how long it says the string is, and no diagram with the problem :(
Careful: The problem says "A light string of length 2R is attached to the surface"

Draw your own diagram! (Just for yourself.)
 
  • #11
I just don't see it...I drew the diagram, but I don't see what I'm trying to find...
 
  • #12
grouchy said:
I just don't see it...I drew the diagram, but I don't see what I'm trying to find...
Your diagram should show a sphere hanging from a string attached to the ceiling. The string is length 2R and attaches to the surface of the sphere. Where's the center of the sphere?
 
  • #13
the center would be another distance R. so

2/5mr^2 + 3mL^2 = 17/5mr^2 which is still wrong :(
 
  • #14
ah, since the length is 3R it would be M(3R)^2 so it would be 9MR^2 so..

2/5mr^2 + 9mr^2 = 47/5mr^2 would that be the proper way to do it?
 
Last edited:
  • #15
For the second one...since it is rotating on the far left mass, I can disregard it since the inertia would be of the masses rotating around it right? So...

the middle mass mr^2 = m(L/2)^2 = 1/4mL^2
the far right mass mr^2 where r would be the length to the point of rotation so mL^2

1/4mL^2 + mL^2 = 5/4mL^2
I think I did the two problems the correct way, if there is anything you see shady about my method I'd love to know. THX for your patience and help lol. I think I'm starting to understand this some. Thx again! Peace.
 
  • #16
Your latest solutions to both problems are perfectly correct. Well done!
 

Related to Find the rotational inertia of a sphere suspended from the ceiling

1. What is rotational inertia?

Rotational inertia, also known as moment of inertia, is a measure of an object's resistance to changes in its rotational motion. It is similar to mass in linear motion, but for rotational motion instead.

2. How is rotational inertia different from mass?

While mass is a measure of an object's resistance to changes in linear motion, rotational inertia is a measure of an object's resistance to changes in rotational motion. Mass is a scalar quantity, while rotational inertia is a tensor quantity.

3. How do you calculate the rotational inertia of a sphere suspended from the ceiling?

The rotational inertia of a sphere suspended from the ceiling can be calculated using the formula I = 2/5 * m * r^2, where I is the rotational inertia, m is the mass of the sphere, and r is the radius of the sphere. This formula assumes that the sphere is a solid, homogeneous object.

4. What factors affect the rotational inertia of a sphere suspended from the ceiling?

The rotational inertia of a sphere suspended from the ceiling is affected by its mass and the distribution of that mass relative to the axis of rotation. A sphere with a larger mass or a larger radius will have a higher rotational inertia.

5. How does the rotational inertia of a sphere suspended from the ceiling impact its motion?

The rotational inertia of a sphere suspended from the ceiling determines how easily it can be rotated and how much torque is needed to change its rotational motion. A sphere with a higher rotational inertia will require more torque to change its rotation compared to a sphere with a lower rotational inertia.

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