# Homework Help: Torque and angular momentum with ball shot into air

1. Nov 21, 2005

### physicsklutz

A 0.380 kg ball is shot directly upward with an initial speed of 38.5 m/s. What is its angular momentum about point P, at a horizontal distance of 1.85 m from the launch point, when the ball is at its max height? Halfway back to the ground? What is the torque on the ball about point P due to the gravitational force when the ball is at its max height? Halfway back to the ground?

Where should I begin?

2. Nov 21, 2005

### siddharth

The angular momentum of a particle wrt to a point would be
$$l=r \times p$$.
So, what do you need to find first to find the angular momentum?

3. Nov 22, 2005

### physicsklutz

to find angular momentum when the object is halfway back to the ground, should i use the velocity at that point? and to find r, would i use the hypotenuse of the height of the object and the distance the object is from P?

4. Jan 11, 2008

### Predator

Sorry to resurrect this thread, but I have this exact same question on a problem due tonight. Only difference is my initial velocity is 41.0m/s and point P is 2.05m from the base of the launch point and the mass is 0.360kg.

I used kinematics to find the maximum height:
v^2 - v0^2 = 2*a*D
0 - 41^2 = 2 * (-9.8) * D
D = 85.7m

I'm solving the torque one first.

Torque = rxF = r * F * sin theta

F in this case, in the problem they are defining it as the gravitational force which would be mg = 0.360 * 9.8 = 3.52N

Now, we need to find r.

That would be pythagorean theorem...sqrt ( 2.05^2 + 85.7^2 ) = 85.789m..

So we have mg * r = 0.360 * 9.8 * 85.789 * sin theta

I keep getting the problem wrong because I think I am getting the wrong value of theta?

Theta is the tan inverse of (2.05 / 85.7), right?

Here's an image:

5. Jun 9, 2011

### Levanie2000

I believe the angle should be tan^(-1)(y/x) so its your max height over your distance to point p

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