Solving the Rod's Angular Acceleration Problem: Why and How?

[sriram_15_93]

1. The problem:
A Force F is provided at the top of a rod placed vertically placed on a smooth frictionless surface. The length of the rod is L and its mass, M. Find the angular acceleration of the rod at the instant the force is applied.
- Please tell me how to solve this and why you have solved it in that way.

2. I solved this problem about the centre of mass, but can it be solved about any other point?

3. If I can solve it about any point I wish, why so?

4. Is it necessary that a body such as the one in the above problem must rotate about its centre of mass only? How can we state this? What is the proof? How do we find out about which point the problem must be solved.

5. I don't know if my method of solving was correct, but when I solved it about the centre of mass I got one answer, and when i solved it about the point in contact with the ground, I got a different answer. So, is angular acceleration different about different points? Or is angular acceleration an intrinsic property? If so, why? Are the answers I get (by solving in the aforementioned two ways) the angular acceleration relative to the respective points about which I have solved the question? If they are, how do I get the answer in the ground frame? What would have been the pseudo forces if it was relative acceleration I was calculating?

I solved the problem in the following ways:

a) About the centre of mass:
F*(L/2)=I*angular acceleration=(M*L^2)/12*angular acceleration
=>angular acceleration=6*F/(M*L)

b) About the point at the bottom of the rod, in contact with the frictionless horizontal surface:
F*(L)=I*angular acceleration=(M*L^2)/3*angular acceleration
=>angular acceleration=3*F/(M*L)

 

[tiny-tim]

Welcome to PF!

Hi sriram_15_93! Welcome to PF!

(have an omega: ω and an alpha: α and try using the X² tag just above the Reply box )

A Force F is provided at the top of a rod placed vertically placed on a smooth frictionless surface. The length of the rod is L and its mass, M. Find the angular acceleration of the rod at the instant the force is applied.
…
I solved the problem in the following ways:

a) About the centre of mass:
F*(L/2)=I*angular acceleration=(M*L^2)/12*angular acceleration
=>angular acceleration=6*F/(M*L)

b) About the point at the bottom of the rod, in contact with the frictionless horizontal surface:
F*(L)=I*angular acceleration=(M*L^2)/3*angular acceleration
=>angular acceleration=3*F/(M*L)

a) is correct.

In b), you've used the moment of inertia about a point other than the c.o.m. or the centre of rotation, and you can't do that, you have to use the moment of inertia about the c.o.m. to get L_spin, and then add on the L_orbital as an extra, to get the total L.

Torque = d/dt (L) = d/dt (L_spin + L_orbital)

= d/dt (I_c.o.m.ω + r_c.o.m.x v_c.o.m.) = I_c.o.m.α + r_c.o.m.x a_c.o.m.

"Torque about point P = I_Pα" (using I_P as the moment of inertia about the point P) only works if P is the c.o.m. or the centre of rotation.

(and it works for the centre of rotation, O, only because I_c.o.m.α + r_c.o.m.x a_c.o.m. = I_c.o.m.α + r_c.o.m.x (r_c.o.m.x α_c.o.m.) = I_c.o.m.α + r_c.o.m.²α = I_Oα by the parallel axis theorem.)

In this case (leaving out all the Fs and Ls …), a_c.o.m. = 1, and α = 6, and the torque about a point distant r from the force is r.

If you take moments about the centre of the rod (r = 1/2), dL/dt = 6*1/12 = 1/2 = r

If you take moments about the top of the rod (r = 0), dL/dt = 6*1/12 minus 1*1/2 = 0 = r

If you take moments about the centre of rotation (r = 2/3), dL/dt = 6*1/12 plus 1*1/6 = 2/3 = r (or dL/dt = 6*1/9 = 2/3 = r)

If you take moments about the bottom of the rod (r = 1), dL/dt = 6*1/12 plus 1*1/2 = 1 = r ​

 

[sriram_15_93]

Thanks a lot Tiny-Tim! So, spin angular momentum an intrinsic property while orbital angular momentum depends on the point about which we are calculating torque, is it? And about the signs (6*1/12 plus 1/2 or 6*1/12minus 1/2) - how did you get them? I've always been solving sums only about the c.o.m because I've been told it simplifies calculations but I've never known why. So the concept of spin angular momentum and orbital angular momentum is new to me. So please explain how you got a plus or a minus in the equations.
and another doubt:
By r_c.o.m I think you mean the distance of the reference point from the centre of mass. but does a_c.o.m mean the acceleration of the centre of mass, or the acceleration of the reference point w.r.t the c.o.m?
Thanks once again!

 

[tiny-tim]

So, spin angular momentum an intrinsic property while orbital angular momentum depends on the point about which we are calculating torque, is it?

Exactly.

And about the signs (6*1/12 plus 1/2 or 6*1/12minus 1/2) - how did you get them?

I just look at the diagram to see whether the c.o.m. is going clockwise or anti-clockwise.

The c.o.m. in this case moves to the right, so from above it's anti-clockwise, and from below it's clockwise (and the spin is clockwise).

By r_c.o.m I think you mean the distance of the reference point from the centre of mass. but does a_c.o.m mean the acceleration of the centre of mass, or the acceleration of the reference point w.r.t the c.o.m?

I meant the position or acceleration of the c.o.m from my "origin" (and I would always have a_c.o.m = r''_c.o.m anyway).