Torque and angular momentum of a conical pendulum

[joe_coolish]

Homework Statement

A ball (mass m = 250 g) on the end of an ideal string is moving in circular motion as a conical pendulum as in the figure. The length L of the string is 1.85 m and the angle with the vertical is 37°.

1. What is the magnitude of the torque exerted on the ball about the support point (Nm)?
2. What is the magnitude of the angular momentum of the ball about the support point (Kg m^2/s^2)?
3. What is the direction of the angular momentum of the ball about the support point?

Homework Equations

t= r x F
t= rFsinθ
L= r x p
p= mv
v= 2PI * r * r/period
period= 2PI*SQRT(h/g)

T=mg/cosθ

The Attempt at a Solution

1) My first attempt at finding the torque was to find the the tension of the string (T) which equalled:

T=3.0677

I figured that T is the only force acting on support point so I took T and plugged it into

t= rFsinθ

and got 2.055. The correct answer was 2.73. After playing with the numbers, switching sin with cos got me the correct answer

t= rFcosθ = 2.7277

I don't know if that is coincidence or what. Is that correct?

2) I took F and replaced it with p and got 0.637. The correct answer was 1.33, which is why I think that the formula to get the answer for 1) was just coincidence. I'm not too sure where to go here.

3) I have no idea!

 

[Doc Al]

1) My first attempt at finding the torque was to find the the tension of the string (T) which equalled:

T=3.0677

I figured that T is the only force acting on support point so I took T and plugged it into

t= rFsinθ

and got 2.055. The correct answer was 2.73. After playing with the numbers, switching sin with cos got me the correct answer

t= rFcosθ = 2.7277

I don't know if that is coincidence or what. Is that correct?

Not sure what you mean when you say you 'played with the numbers'. In any case, you're finding the torque on the ball about the support point. (The tension is parallel to r, so it contributes no torque directly.) What's the net force on the ball?

 

[joe_coolish]

Not sure what you mean when you say you 'played with the numbers'. In any case, you're finding the torque on the ball about the support point. (The tension is parallel to r, so it contributes no torque directly.) What's the net force on the ball?

Thank you for the reply!

All of the forces that are acting on the ball are from gravity and the tension of the string. As well as the centripital force pushing the ball inward. Am I missing another force?

As for "playing with the numbers" I meant I didn't know what to do so I changed the sin to cos to see if I had made a mistake in my formula. Those kinds of things.

 

[Doc Al]

All of the forces that are acting on the ball are from gravity and the tension of the string.

Right. So calculate the torque from each.

As well as the centripital force pushing the ball inward.

Careful! It's the tension of the string that provides the 'centripetal force'. Centripetal force is not a separate force of its own.

Am I missing another force?

Nope. Just two.

As for "playing with the numbers" I meant I didn't know what to do so I changed the sin to cos to see if I had made a mistake in my formula. Those kinds of things.

Sure. Now try it again for real. Calculate the torque using the formula T = r X F.

 

[joe_coolish]

... Calculate the torque using the formula T = r X F.

ok, this is where my brain stops working. r and F are both vectors (correct?), and I'm assuming that r is the position vector of the ball, and F is the force vector of the 2 forces acting on the ball?

if so, would r= Lsinθi + Lcosθj + (?)k and F= -mgi + Tj + (?)k? How do I find the k values? or are they not important?

 

[Doc Al]

r and F are both vectors (correct?), and I'm assuming that r is the position vector of the ball, and F is the force vector of the 2 forces acting on the ball?

Yes. You can also calculate the torque due to each force separately, then add them up. (That's what I would do.)

if so, would r= Lsinθi + Lcosθj + (?)k and F= -mgi + Tj + (?)k?

You could do it that way if you like (but careful with how you define your coordinates; also, you'll need the components of the tension). I think it's easier to just use rXF = rFsin(theta). Note that the angle in that formula is the angle between r and F.

How do I find the k values? or are they not important?

Two coordinates are enough for defining r and F.

 

[joe_coolish]

Ah! That works :)

r(mg)sinθ = 2.73

and since the tension is θ = 0; that would be 2.73 + 0 = 2.73 :)

and for Angular Momentum, it would be rxp = r(mv)sinθ? My calculation for v must be wrong. Am I using the right formula to calculate v?

 

[Doc Al]

Ah! That works :)

r(mg)sinθ = 2.73

and since the tension is θ = 0; that would be 2.73 + 0 = 2.73 :)

Good!

and for Angular Momentum, it would be rxp = r(mv)sinθ?

Good. (Careful with θ, which is the angle between r and v.)

My calculation for v must be wrong. Am I using the right formula to calculate v?

What formula are you using? (Just apply Newton's 2nd law.)