Rotational Kinematics - Torque

1. Jan 9, 2010

nahanksh

1. The problem statement, all variables and given/known data

https://online-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys211/spring09/homework/09/pivoting_stick/5.gif [Broken]
A stick of uniform density with mass M = 8.7 kg and length L = 0.6 m is pivoted about an axle which is perpendicular to its length and located 0.14 m from one end. Ignore any friction between the stick and the axle.
The stick is held horizontal and then released.
What is its angular acceleration as it passes through the vertical position?

2. Relevant equations
Torque = I*(angular acceleration)

Torque = r X F

3. The attempt at a solution

I need to find the net torque of the system to solve this problem.
But when i draw a free-body diagram, i can't find the tangential force to the stick which should be used in the torque equation.(I find only horizontal forces which make no sense...)

Thanks.

Last edited by a moderator: May 4, 2017
2. Jan 9, 2010

Staff: Mentor

What forces act on the stick? It seems you are leaving out an obvious one.

3. Jan 9, 2010

nahanksh

You mean when it's vertical?
I think you meant the "gravity" i guess..?
Otherwise,Could you explain..?

4. Jan 9, 2010

Staff: Mentor

Yes, gravity is what I was thinking of. But I misread the problem... I thought it was let go from the vertical position. D'oh!

In any case, do any horizontal forces act on the stick (forces that can exert a torque, that is) when it is vertical? What's the torque on the stick at that point?

5. Jan 9, 2010

nahanksh

No horizontal forces when it's vertical, i suppose?

Then there is no torque at that point, hence, angular acceleration is zero..

I thought this was a tricky question.. never considered this would be the answer...

Thanks a lot !!!!

6. Jan 9, 2010

nahanksh

The question is
"What is the magnitude of the vertical component of the force exerted by the stick on the axle when the stick passes through the vertical? "

And i have got all the numerical things.
I thought the vertical component should be "ma - mg" at vertical position. (here, 'a' is centripetal acceleration)

But it turns out that it should be "ma + mg" must be the vertical component..

I think centripetal acceleration points to the pivot(+y direction) and mg points down(-y direction) at vertical position..
Then, how come the ADDITION happens here...?

Thanks a lot.

7. Jan 9, 2010

Staff: Mentor

Right.
Just apply Newton's 2nd law.

In general: ΣF = ma

Applying that to this particular problem: Fp - mg = +ma
Thus: Fp = mg + ma

Make sense?

8. Jan 9, 2010

nahanksh

It totally makes sense...!!
Thanks a lot !!

God bless you !