# Falling/rotating rod problem

1. Nov 24, 2006

### curt2121

a 1kg rod of unknown length has a spherical base with a mass of 2kg and a mass of 1kg on the top. if it is allowed to fall over from vertical, what will the velocity of the 1kg mass at the top be when it falls down (no friction). a friend is driving me crazy with this since i gave him a little brain teaser before but physics isn't my thing so i'm not sure how to go about this. i can make a diagram later if that helps. thanks

2. Nov 24, 2006

### Danger

I might be missing something here, but I don't think that the question is answerable as stated. I don't see how you can calculate the terminal speed of the top mass if you don't know how long the connecting rod is.

3. Nov 24, 2006

### curt2121

well the speed would be in terms of the length i'm assuming

4. Nov 24, 2006

### Staff: Mentor

Consider energy conservation.

5. Nov 25, 2006

### curt2121

um not really sure why this was moved since it isn't for a class but ok I guess. anyone knw of a solution?

6. Nov 26, 2006

### Staff: Mentor

This the forum for all textbook-type problems, whether for a class or on your own.

To solve the problem just use conservation of mechanical energy. Hint: If there is no friction, how will the center of mass move? What's the change in gravitational PE? (You may have to make some assumptions as to the size of the masses--treat them as point masses, for instance.)

7. Nov 26, 2006

### curt2121

for some reason i keep wanting to include the moment of inertia for the sphere (.4MR^2) since the base will have to roll over. Am I completely off on that? help is much appreciated since i'll have to take him to lunch wednesday if I don't figure it out

8. Nov 27, 2006

### Staff: Mentor

Not at all. Include the moments of inertia of all the parts.

9. Nov 27, 2006

### curt2121

ok i'm not quite sure how they're all supposed to fit together

10. Nov 27, 2006

### samo267

lol the mass won't give u any velocities as the mass doesn't it the only things that affect velocity are acceleration, distance and time since u dont have distance nor time how r u goin to solve it
PS if u take him to lunch ask him 4 the answer and u'll find that there's a trick u didnt get it in u answer just like a cock has to get out an egg it's standing on a wall where the right of it there's fire and the left there's water where does it hav to put the egg??

11. Nov 27, 2006

### Staff: Mentor

If you have the sizes of the pieces (in terms of the length of the rod, perhaps) you can calculate the total moment of inertia of the system (which is just the sum of the parts). You'll need to relate the rotation of the system to the motion of the center of mass. (Which may not be so easy, since the base is spherical.)

If you were willing to model it as a stick with point masses on each end, the problem would be easier to solve.

12. Nov 27, 2006

### curt2121

from looking at the problem, i think it's safe to model the top mass as a point since it is fixed, but the bottom has to be spherical, that's the whole trick to it. I'm attaching a diagram for those visual learners. i know (at least i think if I remember right) the center of mass is:

cm of l=.5l (1kg)
cm of M(big mass at bottom) we'll say is at 0 and is 2kg
cm of m (little mass at top) is l and is 1kg
so the cm for the system is:
[(0*2)+(.5l*1)+(l*1)]/(4kg)= .375l
so that the cm of the system is 3/8 of the way up the length of the rod. am i good so far?

#### Attached Files:

• ###### diagram.JPG
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13. Nov 28, 2006

### curt2121

am i on track with that cm? and i'm not quite sure how to combine the moments of inertia of the different objects

14. Nov 28, 2006

### Staff: Mentor

Sure. Given your assumptions (spheres centered at y=0 and y=L; stick centered at L/2), your center of mass calculation is good.

To calculate the moment of inertia, you'll need to go a step further in assumptions. How big is the bottom sphere compared to the stick? For example: If the sphere has radius L/2, then the stick is only an L/2 long projection from the sphere. Treat the system as a set of three non-overlapping objects, find the rotational inertia of each object about the cm of the system and add them up.

(I hope you agreed upon a complete description of the problem before you accepted the challenge!)

15. Nov 28, 2006

### curt2121

so that the total moment of inertia is 2/5*2*R(radius of large mass)^2 +2/5*1*r(radius of small mass)^2, + 1/12*1*L^2=

4/5R^2+2/5r^2+1/12L^2?

then assuming the top and bottom masses have the same density, the volume of the top is half the volume of the bottom so that the radius of the larger object is 2^(1/3) times the radius of the larger object so that

R=r*2^(1.3) and the total equation becomes:

4/5r^(2/3)+2/5r^2+1/12L^2

does that seem right? did it get me closer?

16. Nov 28, 2006

### Staff: Mentor

Those are the moments of inertia of each piece about its own center of mass.

So you can't just add them. Use the parallel axis theorem to find the moment of inertia of each object about the cm of the entire system. Add those up and you'll have a meaningful moment of inertia for the system.

Also: If the stick is L long, do you mean for the complete object to have an overall length of: 2R + L + 2r ?

17. Nov 28, 2006

### curt2121

that last part abotu the length is correct. i don't remember ever hearing anythign about the parallel axis problem before. Well, it's possible i learned it as a different name. can you please enlighten me?

edit: just looked it up on wikipedia but I definitely don't understand the explanation there. it says

Iz=Icm+Md^2

I'm trying to solve for ICM so i want Iz-Md^2 but that just confuses me

Last edited: Nov 28, 2006
18. Nov 28, 2006

### curt2121

now that i'm thinking, i hope i explained the problem right from the beginning. the rod isn't rotating about itself, ir's simply on a surface, say a desk and allowed to plop over to horizontal. does all this adding the moments of inertia still apply?

19. Nov 29, 2006

### OlderDan

Yes it applies. The system will rotate as it falls, so some of the kinetic energy will be associated with the angular motion.

20. Nov 29, 2006

### Staff: Mentor

Here's a useful discussion of the http://hyperphysics.phy-astr.gsu.edu/hbase/parax.html#pax". Briefly, if you know the moment of inertia of an object about an axis through its center of mass, the parallel axis theorem allows you to calculate its moment of inertia about any parallel axis. For example: You know that the moment of inertia of a stick about its center is 1/12 M L^2. You can use the parallel axis theorem to find its moment of inertia about the end:
$$1/12 M L^2 + M D^2 = 1/12 M L^2 + M (L/2)^2 = 1/3 M L^2$$

No, you have Icm and need Iz.

In your case you have the moments of inertia of each piece about its own center of mass. So you can use the theorem to compute the moments of inertia about the center of mass of the entire system.

Last edited by a moderator: Apr 22, 2017
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