Circular Path of a Pendulum Problem

[whoopie88]

Homework Statement

The diagram shows an object of mass 3.0 kg traveling in a circular path of radius 1.2 m while suspended by a piece of string of length 1.9 m. What is the centripetal force on the mass?

Diagram:

Homework Equations

All circular motion formulas.

The Attempt at a Solution

Tx = Fc
1.9sin39 = Fc
1.2 N = Fc

 

[turin]

1.9sin39 = Fc

Where did this come from? I suggest expressing your equations genericly first, so that we can determine what you mean. All that I can do is guess that the "1.9" refers to the length of the string and the T and F are forces, so that you are trying to determine a force from a length. Does that make sense?

What physical principle indicates that there should be a centripetal force on the mass?

 

[whoopie88]

I'm not sure how else to do it. Can you help me?

 

[turin]

Explain, physically, what is indicated by the diagram. What object is being acted on by forces, and what forces? Can you determine a relationship among the forces?

 

[whoopie88]

Well it looks like there's an object suspended by a string, and it's swinging around in a circle. From the string, there's a Force of Tension, and there's also the Force of Gravity acting on the ball. And there's a centripetal force towards the center of the circle.

As for a relationship between the forces...I'm not sure. If I use the sum of the forces in the x-direction, it would be Fnet = Fc. If I did the sum of the forces in the y-direction, it would be Fnet = T - Fg.

Would it be appropriate to say Fc = T - Fg? (I'm not sure how to find the Force of Tension, either)

 

[turin]

OK, excellent. Now we have something to work with.

From the string, there's a Force of Tension, and there's also the Force of Gravity acting on the ball. And there's a centripetal force towards the center of the circle.

Great. I concur with this. However, you should be more specific. Distinguish between applied forces and reaction forces. That is, in Newton's second law, there are the F forces and the m.a force. Two of the forces that you mentioned are F forces. That is, they are applied due to the contact or presense of certain objects. Then, there is the resulting m.a force, which is manifested in the motion as a.

If I use the sum of the forces in the x-direction, it would be Fnet = Fc. If I did the sum of the forces in the y-direction, it would be Fnet = T - Fg.

Not quite. Is the tension exactly vertical? Have you drawn a free-body diagram? Hint: is there any vertical acceleration?

Would it be appropriate to say Fc = T - Fg?

No. But, there may be some relationship involving these forces. Hint: I see that you are familiar with trigonometry.

(I'm not sure how to find the Force of Tension, either)

You may not necessarily need to do this. Try to work it out symbolically and see how far you get. This is usually daunting for the novice, but after a few of these problems you will realize how much better you can sniff out the solution, because it forces you to think in terms of the physical relationships rather than the numerical relationships. As a bonus, it will help you practice your algebra, and it will save you a lot of extra calculational work in the long run.

 

[whoopie88]

Great. I concur with this. However, you should be more specific. Distinguish between applied forces and reaction forces. That is, in Newton's second law, there are the F forces and the m.a force. Two of the forces that you mentioned are F forces. That is, they are applied due to the contact or presense of certain objects. Then, there is the resulting m.a force, which is manifested in the motion as a.

The Force of Tension and the Force of Gravity are applied forces, correct? So the centripetal force is the reaction force.

Not quite. Is the tension exactly vertical? Have you drawn a free-body diagram? Hint: is there any vertical acceleration?

So Fnet = Ty - Fg, correct? (for the forces in the y-direction) And Fnet = Tx + Fc for the forces in the x-direction.

 

[turin]

You are on the right track. Can you figure out the relationship between Tx and Ty? Also make sure to address the issue of the vertical motion (physically intuitively).

 

[whoopie88]

This is where I'm lost. I can't find a relationship between Tx and Ty without using trigonometry, which is what I tried doing in the attempt at a solution in the first post.

Also, I'm not sure about the vertical motion. It doesn't look to me like there is any vertical motion. Am I wrong?

 

[bozobuddy]

3*9.8=29.4N
29.4N tan 39 = fx
done...