# Angular momentum of ball spinning on a pendulum around a support point

1. Mar 22, 2012

### Los Frijoles

1. The problem statement, all variables and given/known data
A ball (mass M: 0.250kg) is attached to a rope (length l: 1.69m) that is fixed to a point somewhere. The ball is spinning in a circle around the support point so that the rope makes an angle 37 degrees from the vertical. I have to find the magnitude of the angular momentum of the ball about the support point, which I can't figure out for the life of me.

2. Relevant equations
r = sin37l = 1.02m
F = Mg = 0.250kg * 9.8m/s^2
t (torque) = r x F = rFsin90 = 1.02m(.250)(9.8) = 2.49 Nm
..as for getting the angular momentum all I know is:
L = r x p = mvr (in this situation)

3. The attempt at a solution
Part of the question was to get the torque, which after getting it wrong once and being shown the answer (and then given different data...that's how the computer grading this works) I figured out that where I was previously doing t = rFsin37 that it needed to be rFsin90 since it is from the perspective of the anchor point. After trying to reverse engineer the answer I got back for the angular momentum I have come up with nothing and I have no idea how to figure it out.

I think the problem lies in that I need to somehow transform torque into the tangential velocity v required for the equation for L and I can't find anywhere in my textbook where it says that and I can't seem to find any relating equations except t = dL/dt (and I don't have any dt (and I haven't yet completed my calculus course...yay :/...))

Can anyone help me figure this out?

2. Mar 22, 2012

### emailanmol

Hey,

This is a classic problem.

First thing you need to realise is that angular momentum and Torque are dependent on the axis (or point of calculation).

If you calculate the torque (or angular momentum) with respect to the centre of the circle and torque (or angular momentum)with respect to point of suspension you will get two different answers.

Net Torque of a particle about a point is given by R X F ,
where F is the net force acting on the particle ,(same with respect to all points as value of net force vector is frame independent ), and r is the position vector of the particle wrt that point.

The net force acting on the particle is it's mass times centripetal acceleration.
What is it?

What is its direction.?

What is the direction of R for the body?

(NOTE:you are wrong in considering net force as mg.Its something else and obviously dependent on 37 degrees).

the angular momentum about a point for a particle is defined again as m(R X v).

Remember R here is position vector of particle wrt that point, and v is its relative velocity wrt that point.

Now imagine the figure,

The relative velocity of the particle wrt point of suspension is tangential to the circle.

(WHY?)

What you need to realise now is that The vector R (which is along the string) is actually perpendicular to this velocity.

It is very important to realise that R is perpendicular to V.

One way to do it is to take components of R .

Take one component along the line joining the point of suspension and centre of the circle.
The other will now be along the line joining the particle and the centre of circle.

Both these components are perpendicular to the velocity, so R is perpendicular to velocity.

Use Formula to find magnitude.

A better way is to write everything in the form of vectors.

Take the line joining the centre of circle and suspension as one axis.

Take the other two axis along the radius of circle and tangential to it. These two axis move along with particle.

Write R, V and F in terms of vector and apply vector cross product. The advantage of this that it also gives you direction of the quantities in terms of vectors.

Last edited: Mar 22, 2012