# Homework Help: Centripetel acceleration and tension

1. Nov 15, 2005

### dnt

the question is a 2.0 kg object is attached to a 1.5 m long string and swung in a vertical circle at a constant speed of 12 m/s.

a. what is the tension in the string when the object is at the bottom of its path

b. what is the tension in the string when the object is at the top of its path

can someone help me set up the free body diagram for both. what i think is that at the bottom, you have weight down, centripetel force down, and tension up? is that right?

but after that, i still dont konw how to set it up because net force equals ma, but what is the acceleration?

2. Nov 15, 2005

### Päällikkö

At the bottom, you have two forces: tension of the string and gravity.
Sum of these should equal centripetal force (which is, of course, upwards).

3. Nov 15, 2005

### dnt

then at the top you have gravity and centripetel force down which equals tension (pointing up?) is that correct?

general question: how do you know which way tension points in free body diagrams with strings?

4. Nov 15, 2005

### andrewchang

tension can only be downwards when it is at the top- tension can never push objects.

5. Nov 15, 2005

### dnt

dont these two statements contradict each other?

still not understanding how to determine the direction of tension, especially in a question like this?

6. Nov 16, 2005

### Päällikkö

No. I wrote about tension at the bottom, andrewchang posted about the top of the loop. The magnitude and direction of tension changes during the loop.

Tension points to the center of the cirle (along the rope).

7. Nov 16, 2005

### dnt

then im still confused. you said at the bottom, tension and gravity should equal centripetel force.

well, if gravity points down, and tension points up, how can they add up to equal centripetel force (which is up)?

i drew a free body diagram, and at the bottom i have gravity (down), centripetel (up) and tension (up).

shouldnt centripetel + tension = gravity?

8. Nov 16, 2005

### Päällikkö

T + (-G) = Fc
Gravity has a negative sign (or tension, depending on how you choose the positive directions. Anyways, they have different signs).

9. Nov 16, 2005

### dnt

ok then my problem arises from relating the free body diagram to the net force equation. i was taught that all the forces in one direction minus the forces in another direction equals mass times acceleration. well in the up/down direction, a=0, so all the forces pointing up should equal all the forces pointing down. is that logic correct?

well if so, and you said gravity is down, centripetel force is up and tension is up (again, this is at the bottom of the swinging objects path), shouldnt it be:

Fc + T = G

but you said: T + (-G) = Fc which is the same as Fc + G = T

basically the T and G are switched.

im so confused.

10. Nov 16, 2005

### Päällikkö

a is clearly not 0, the particle undergoes circular motion!

Fc isn't a force, it's the sum of forces.

The two forces acting on the particle are tension and gravity (as stated before):
T - G = ma = mac = Fc

11. Nov 16, 2005

### dnt

oh i was looking at it in the vertical direction and assumed a=0 (thought the acceleration was in the horizontal).

now that equation makes sense. thanks.

for the top i could say T + G = Fc? is that right? or is it -T - G = Fc?

12. Nov 16, 2005

### Päällikkö

There is no acceleration in the horizontal, as there's no forces acting horizontally (this changes at the instant the rope isn't perfectly vertical, though).

T + G = Fc is good, as it's the same as:
-T - G = -Fc (the same choice of positive/negative directions)

Last edited: Nov 16, 2005