# Newton's 2nd Law for a Particle in Uniform Circular motion

1. Jun 24, 2008

### clintau

1. A 4.0kg object is attached to a vertical rod by 2 strings. Object roatates in a horizontal circle at a constand speed 6.00 m/s. Find the tention in the upper and lower string. The strings are 2 meters long. the distance between the attachment points for each string is 3 meters.

2. F=ma=m(v^2/r)

3. I know that you must sum the forces in the x and y direction but cant seem to get rolling on this

2. Jun 24, 2008

### G01

Yes, you must sum the forces from both strings. The sum of the forces is a net centripetal force. What does this mean it has to be equal to?

HINT: What are all centripetal forces equal to?

3. Jun 24, 2008

### clintau

i don't know.

4. Jun 24, 2008

### rock.freak667

Draw a vector diagram...then split the tension into it's components.

5. Jun 24, 2008

### clintau

Y direction
Fupper= T*cos(theta)+mg=0
Flower= T*cos(theta)-mg=0

yes no?

6. Jun 24, 2008

### G01

Yes, you do. You said it in your first post.

F_c=mv^2/r

So, if:

$$\Sigma F_{radial direction} = F_c$$ and $$F_c=\frac{mv^2}{r}$$ then...

Also,

You need to sum the radial components (x components) of the forces, since only those will contribute to the net centripetal force. To find these it would help to draw a free body diagram as clintau said. How far can you get now?

7. Jun 25, 2008

### clintau

so Fc=4*6^2/r=109.09

That gives the force in the x direction....What about the Y?

8. Jun 25, 2008

### clintau

how would you find the tensions in the strings?

9. Jun 25, 2008

### rock.freak667

$$\Sigma F_{vertical}=0$$

10. Jun 25, 2008

### clintau

Fcx is the T in the upper string. How do you get the T in the lower?

Tupper-mg? gives app 69N. Answer should be 56.2

what is wrong?

11. Jun 26, 2008

### rl.bhat

Instead of taking componets of tention on Y direction, take the component of mg along the tensions.
So T(upper) = T1 + mgcos(theta)
T(lower) = T2 - mgsin(theta)
Now T(upper), T(lower) and Fc are in equilibrium. Apply the Lami's theorem.
[T1 + mgcos(theta)]/sin(pi/2 + theta) = [T2 - mgsin(theta)]/sin(pi/2 + theta) = [mV^2/r]/sin(pi -2*theta).
Now solve for T1 and T2.