# Homework Help: Radial force of a rotating stick

1. Nov 14, 2008

### bollocks748

Radial force of a rotating stick!!!

1. The Problem
In this problem we want to learn a little bit about what is sometimes called dynamical loading. Our simple system consists of a uniform stick of length L and mass M hinged at one end. We would like to calculate the forces on the (frictionless) hinge when the stick is released from rest at an angle theta_0 with respect to the vertical. You may find it useful to combine work and energy equations with torque (N II) equations.

3.1 Show that the radial force exerted on the stick by the hinge is F_r = Mg/2*(5 cos(theta) - 3 cos(theta_0)), where theta is the angle of the stick with respect to the vertical after it is released.

3.2 Show that the tangential (tangent to the direction of motion, perpendicular to the stick) force exerted on the stick by the hinge is F_t = Mg/4*sin(theta).

2. The attempt at a solution

Alright, I've spent several hours trying to figure this one out, and it's really killing me. I didn't use torque in the solution, but I did set it up initially like this:

t= L/2 * mg sin(theta) = I * alpha

I then set up the energy as the height of the center of mass tilted to theta_0 equal to the height relative to the tilting of theta plus both forms of kinetic energy.

mgcos(theta_0)*(L/2)= mgcos(theta)*(L/2) + 1/2 m v^2 + 1/2 I w^2

substituting V/(L/2) for w, I got 2/3 m v^2 for the total kinetic energy. Using that, I set it up so it was equal to mv^2/(L/2) (the radial force), and I was left with a function that has cos(theta_0) - cos(theta), which is the opposite of what I need.

After trying to figure out something else, I tried making an equation for work and substituting that in for KE instead.

I figured the work to be F_r((L/2)cos(theta)-(L/2)cos(theta_0)), the difference between the heights of the initial positions times the radial force. When substituting that in for 1/2 m v^2 + 1/2 I w^2, I ended up with:

mg(.5cos(theta_0)-.5cos(theta)) = F_r ( 1/2 cos(theta)-1/2cos(theta_0)), which can't be simplified to the equation I need. So I am officially stuck!

Any help would be much appreciated. :-)

2. Nov 14, 2008

### katchum

Re: Radial force of a rotating stick!!!

I have a question:

If we are rotating from the hinch, we use the moment of inertia: 1/3.m.L^2?

So is it true that we have for total kinetic energy: 1/2 m v^2 + 1/2 I w^2 = 1/2 . 1/3.m.L^2.w^2?
Or is it: 1/2 mv^2 + 1/2 . 1/3.m.L^2.w^2 = 7/6 mv^2?

Even so, then:
7/6mv^2 = mgcos(theta_0)*(L/2) - mgcos(theta)*(L/2)
Fr = mgcos(theta) - mv^2/(L/2)

Fr = mgcos(theta) - 2m/L. (6/7m)(mgcos(theta_0)*(L/2) - mgcos(theta)*(L/2))
Fr = mgcos(theta) - (6/7)(mgcos(theta_0-mgcos(theta))
Fr = (13/7)mgcos(theta) - (6/7)(mgcos(theta_0)

Otherwise:

2/3mv^2 = mgcos(theta_0)*(L/2) - mgcos(theta)*(L/2)
Fr = mgcos(theta) - 2m/L. (3/2m)(mgcos(theta_0)*(L/2) - mgcos(theta)*(L/2))
Fr = mgcos(theta) - (3/2)(mgcos(theta_0-mgcos(theta))
Fr = (5/2)mgcos(theta) - (3/2)(mgcos(theta_0)

I just discovered that apparently 1/2mv^2 + 1/2 Iw^2 is the same as 1/2 Iw^2 with the second "I" being the moment inertia with the axis not in the center. So basically this "I" which is not in the center is already taking in account, the velocity of the center point mass of the rod?

For the tangential force problem I get:

M = I alpha = L/2.sint.g.m = 1/3.m.L^2 . alpha

a = alpha . L/2 = 3/4 sint.g
Ft = mg 3/4 sint??

Somehow this looks a bit like the solution to problem 2)
I'm stuck too. Shouldn't the tangential force at 90 ° be mg? I think not because when you have a very long rod and you hold the one end of the rod and let the rod swing down coming from the 90 ° then you won't feel the gravitational mass of the whole rod in your hand. Don't you think?

Last edited: Nov 14, 2008
3. Nov 14, 2008

### bollocks748

Re: Radial force of a rotating stick!!!

I am equally confused :). But thanks a lot for the help on the first part!

If you treat the torque arm to be the full length of the stick, you get 3/2mgsin(theta) to be the torque. The only thing that comes close to equaling 1/4mgsin(theta) is if you treat the initial angle as not part of the torque and set the equation up like so:

3/4mgsin(theta)-1/2mgsin(theta_0)

Which subtracts the way I initially set up the torque. This comes SO close to the setup I want, but I don't think I can resolve the two different thetas... so I'm about 80% sure this won't work. I would think that at 90, the value would be mg... but there must be some other force taking away from the value. The only other component I can think of is radial force, but that acts perpendicular to the tangential force, and centripetal force is technically nonexistant, so I don't think that works either.

4. Nov 14, 2008

### alphysicist

Re: Radial force of a rotating stick!!!

Hi bollocks748,

There is a lot of information in your and katchum's post, so I apologize if I'm repeating any of it!

This is good, but go ahead and put in the formula for I here.

I don't think this is right. These potential energy terms should be negative (because you're measuring from the pivot down). Also, if you calculate the kinetic energy about the same pivot as your torque equation then (1/2)I w2 will account for the total kinetic energy (and you do not need to the 1/2 mv2 term).

If you now draw a force diagram (with unknown radial and perpendicular hinge forces at one end) and apply Newton's law to find the center of mass linear acceleration (in the radial and tangential directions) you should get the correct answer for both parts. What do you get?