Angular Momentum: Spinning Mass

[JoeyBob]

Homework Statement

See attached

Relevant Equations

L=r x p, Torque=change in L

First I calculated the momentum of m1. Since m2 was at rest after the collision, all its momentum was transferred, so m1 has a momentum of 158 i hat.

L=r x p, so its 916 k hat. This would also be the change in L because it was initially 0 when m1 had no velocity, so I know this is the net torque.

So I know the i hat component of torque is 0, so the normal force is the negative of the gravity force, so the friction force is

F=uFn=umg=-3.4496 i hat

Now Net toque=Torque from friction+Torque from Tension

Torque tension=916/(-5.8 x -3.4496)

=-45.782

Divide this by the radius and I get 7.89 N. The answer is suppose to be 3912.85 N

 

[JoeyBob]

I think I am screwing up when I am converting angular momentum into torque, but I am unsure how. I thought net torque was change in angular momentum. And if m1 had an initial angular momentum of 0, I would assume its final angular momentum is the same as its change in angular momentum.

 

[haruspex]

I'm not understanding how or why you bring torque into this.
What forces act immediately after the collision?
What are the acceleration components immediately after collision?
So how can you pair up the acceleration components with the forces?

I thought net torque was change in angular momentum.

Haven’t we been around this before? A change in angular momentum is still dimensionally angular momentum; it is not a torque. Torque is the rate of change in angular momentum. It’s like the difference between displacement and velocity.

 

[JoeyBob]

I'm not understanding how or why you bring torque into this.
What forces act immediately after the collision?
What are the acceleration components immediately after collision?
So how can you pair up the acceleration components with the forces?

So I use the velocity of m1 to calculate the centripetal acceleration?

So v^2/r=T-Ff? This suppose to be an angular momentum problem though.

 

[haruspex]

So v^2/r=T-Ff?

In what directions are the centripetal acceleration, the tension and the frictional force?

 

[JoeyBob]

If the radius is negative j hat, then the centripetal acceleration is also negative j hat. The tension would then be positive j hat.

The friction force would be negative i hat I think.

 

[haruspex]

If the radius is negative j hat, then the centripetal acceleration is also negative j hat. The tension would then be positive j hat.

The friction force would be negative i hat I think.

So why did you mix the frictional force and tension in the equation for centripetal acceleration?

 

[JoeyBob]

So why did you mix the frictional force and tension in the equation for centripetal acceleration?

Then 0=Fc+T

0=-v^2m/r j hat + T j hat

mv^2/r=T

This gives me the right answer, but what does this have to do with angular momentum?

 

[Steve4Physics]

mv^2/r=T

This gives me the right answer, but what does this have to do with angular momentum?

I'll sneak in and add:

The tension IS the centripetal force (the force acting towards the centre of rotation).
(The friction force acts perpendicular to T so is not part of the centripetal force.)
That's why "mv^2/r=T

The original question-title "Angular Momentum:Spinning Mass" is misleading. The problem has nothing to do with angular momentum or spinning ('orbital' motion is not 'spinning').

However, if there is a further part to the question where you need to calculate how quickly the rotating mass slows, then you can use torque = rate of change of angular momentum. The torque is produced by the friction.

If that helps!

 

[JoeyBob]

I'll sneak in and add:

The tension IS the centripetal force (the force acting towards the centre of rotation).
(The friction force acts perpendicular to T so is not part of the centripetal force.)
That's why "mv^2/r=T

The original question-title "Angular Momentum:Spinning Mass" is misleading. The problem has nothing to do with angular momentum or spinning ('orbital' motion is not 'spinning').

However, if there is a further part to the question where you need to calculate how quickly the rotating mass slows, then you can use torque = rate of change of angular momentum. The torque is produced by the friction.

If that helps!

So is Tension always the centripetal force when something is orbiting around something by a rope?

 

[Steve4Physics]

So is Tension always the centripetal force when something is orbiting around something by a rope?

Definitely not.

The centripetal force is the total force acting towards the centre of rotation.

For example think about a mass on a rope whirled it in a vertical circle.
In this case, the centripetal force is a combination of the tension (T) and weight (W), and the combination changes depending on the position of the mass.
At the top position, the centripetal force is |T| + |W| (because both act down, towards the centre).
At the bottom position centripetal force is |T|- |W| (because tension acts up towards the centre and weight acts down away from the centre).
At other positions, centripetal force is a |T| + the component of weight in the direction T.

You have to understand the meaning of centripetal force and think about each problem individually. Try watching a few YouTube videos on this.