# Rotational Motion Problem - 13

1. Dec 16, 2013

### coldblood

Hi friends,

The problem is as:

https://fbcdn-sphotos-g-a.akamaihd.net/hphotos-ak-frc1/q71/s720x720/1506666_1461728184054287_908841020_n.jpg

Attempt -

https://fbcdn-sphotos-f-a.akamaihd.net/hphotos-ak-ash4/q71/s720x720/1003783_1461728430720929_235821158_n.jpg

2. Dec 16, 2013

### haruspex

Same issue as in your problems 4 and 7.
One trick you can use here is to take moments about the point of contact with the ground. The angular acceleration is independent of the reference point.

3. Dec 17, 2013

### coldblood

I am not getting that why should I do so, even after that,

Taking τ = Iα about point of contact,

Mg(R) = 2(3m)R2 (ring) + m(2R2)(particle)(as distance of particle from bottommost point is R√2) × α

Which gives the answer, α = g/8R

4. Dec 17, 2013

### haruspex

Which I believe is correct. If the book says g/3r it's wrong.

5. Dec 17, 2013

### coldblood

What I got from our discussion is that, Always I have to take M.I. about a point, while using τ = Iα, conservation of angular moments and concepts related to that, which is at rest, because torque about it is zero which conserves angular moment of the system.
Is that right?

6. Dec 17, 2013

### coldblood

Why is it SO that the angular acceleration is independent of the reference point?

7. Dec 17, 2013

### haruspex

Consider a rotating rigid object. At time t it has rotated through angle θ=θ(t). If you inspect any part of the object it will have rotated through that angle, and this will be true from any reference frame that has not itself rotated. Now differentiate twice and you see that the same must be true of both the angular velocity and the angular acceleration.
Btw, I owe you a correction on this one. Your error was not the same as on problems 4 and 7. In this problem, there i no sudden impulse, and we are considering the initial acceleration. So we can pick any reference point - they're all effectively at rest.
The error was you forgot the torque from the friction on the ground. By taking moments about the point of contact with the ground we eliminated that.

8. Dec 17, 2013

### coldblood

Well, If we are taking torque about the bottom point the torque of friction about that point would be zero. https://fbcdn-sphotos-a-a.akamaihd.net/hphotos-ak-ash3/1471759_1462378383989267_111677481_n.jpg
https://fbcdn-sphotos-g-a.akamaihd.net/hphotos-ak-prn2/q71/s720x720/960237_1462378397322599_1553793873_n.jpg

9. Dec 18, 2013

### haruspex

In your first two methods in post #8, f=ma is wrong. What's the total mass?
In the second, having taken f the other way, you should now have a the other way too. So $a = -r\alpha$.

10. Dec 18, 2013

### coldblood

So finally should I accept that, α= g/8R ?

11. Dec 18, 2013

### haruspex

Yes. Did you make those corrections and see what you methods lead to? Looks to me like they all come to that same answer.
I was confident the given answer was wrong because if you ignore the opposing torque from friction then you get g/4r, as you did in the OP. Bringing in an opposing torque must make it less, not more.

12. Dec 19, 2013

### coldblood

Yes I solved both the two approaches and both are giving the same answer α = g/8R

13. Dec 19, 2013

### coldblood

Dear haruspex, Please make me clear by connecting all these:

https://fbcdn-sphotos-d-a.akamaihd.net/hphotos-ak-frc3/q71/s720x720/1511150_1462825700611202_1609703582_n.jpg

14. Dec 19, 2013

### haruspex

Recall that I admitted I was wrong in my first posts on 4 and 13. In 7 there is an impulsive (= abrupt) change in angular momentum. Here we have to choose the reference point carefully even for what happens immediately afterwards.
In 13, there is no abrupt change, and we are only concerned with what happens immediately on release from rest. That means that everywhere, even parts of the bodies, are effectively stationary points and can be used as reference axes. Your original method didn't work because you forgot the torque from friction. (Note that when there is an unknown force that you don't need to determine to answer the question, you can often sidestep it by taking moments about a point on its line of action. You could instead take moments somewhere else and combine that with a linear equation to eliminate the unknown force. The two methods produce the same answer, but the first can be quicker.)
In 4, there is no abrupt change, but we are interested in what happens some time after the 'event', so again we need to be careful. [When we talk of a collision we usually mean that there are large forces that act for a short time. We don't know (and perhaps don't care) how large is the force or how short the time - we only only know the momentum imparted, ∫F.dt.
In 4 there is no collision in this sense. It's all pretty smooth.]
I think I mentioned on some thread that there are essentially two safe choices for taking moments: any fixed point or the mass centre of the system. In 4, (after rereading the problem statement), the midpoint of the rod is the mass centre of the system. It is OK because the linear momentum of the system has no angular moment about the mass centre.

15. Dec 19, 2013

### coldblood

I got it haruspex, I will take care for all these. But If in any problem collision takes place like in Prob. 4 and porb 7, we take it as abrupt change or what should we think?

16. Dec 19, 2013

### haruspex

My recommendation is always to take a fixed point as the reference. It might not always be the quickest, but it's always safe.

17. Dec 19, 2013

### coldblood

From now on I will always try to take a fixed point as the reference, then I watch over the quick approach. Thanks a lot for the help.