# Circular motion help

1. Jul 18, 2006

### garyljc

Circular motion help !!!

Hello ,

Here's my question :
A pendulum bob has mass 0.2kg. It is attached to one end of a light rod of length 2m. The rod is free to rotate in a vertical plane about an axis through theo ther end o. Given that the pendulum swings through 60degrees on either side of the vertical
a.) calculate the speed of the bob at the lowest point of its path
b.)the tension in the rod when the rod makes an angle of 30 degrees with the downward vertical

For part a.)
Using the work-energy principle :
v(square) = 2g
therefore v = 4.43m/s

I have no idea how to do part B
My first attempt for this :
Using work-energy principle (slotting r=2 , m =0.2 , theta=30,u=0) to find the velocity at that position , so that I can find the tension using
Tsin30 - 2g = m.v(square)/r

but i just can't get it =( , could anyone help me with tension equations ?

2. Jul 18, 2006

### Staff: Mentor

Use conservation of mechanical energy. Hint: At the top of the motion, the energy is purely potential.

3. Jul 18, 2006

### garyljc

what do you mean conservation of mechanical energy ? well the question i'm doing doesn't describe to be a complete circle motion .
I got part A correct , but it's just part B that cause problems to me

4. Jul 18, 2006

### Staff: Mentor

Once you've found the speed, use Newton's 2nd law applied to the radial direction to find the tension. The acceleration is centripetal.

5. Jul 18, 2006

### Staff: Mentor

I assume by "work-energy" principle you mean conservation of energy. In any case, that equation for radial forces is not quite right. Hint: What's the component of the weight in the radial direction?

6. Jul 18, 2006

### garyljc

the current work-energy principle i'm learning for circular motion is
v(square) = u(square)-2gr(1-cos(theta))

but they said 30 degree to the downward vertical , it can be also left or right =( This will then affect my theta , and hence the calculations ?

If i take 30 to the left my equation will be something like :
Tcos(theta) - mg = m.v(square)/r ........ is it correct ?

7. Jul 18, 2006

### Staff: Mentor

This is just conservation of energy applied to this particular situation.

Does it really matter? Check and see.

No, it's not. How did you arrive at this equation? (Realize that this equation has nothing to do with the earlier equation.) Analyze the forces acting on the bob. Find their components in the radial direction. Apply Newton's 2nd law.

8. Jul 18, 2006

### garyljc

oh okay , my bad ... i'll think about it now , thanks anyway =-)

9. Jul 18, 2006

### garyljc

hey doc , i still can't figure out , could you just help me out with this one ?

10. Jul 18, 2006

### Staff: Mentor

Just do what I asked in my last post:
Analyze the forces acting on the bob. Find their components in the radial direction. Apply Newton's 2nd law.​

11. Jul 18, 2006

### garyljc

doc ,
i've tried to analyse the forces acting . this is what i got (though it doesnt fit the answer )

1. find v(sqaure) , by inserting (theta) as 30 , using the work energy principle
2. then solve the equation Tcos30-0.2g = m.v(square)/r

r=2m , m=0.2

could you see what's wrong with my equations ?

12. Jul 18, 2006

### Staff: Mentor

That equation is not right. Do this:
- List the forces that act on the bob (there are only two)
- Find the components of those forces in the radial direction (parallel to the rod)
- Set the sum of those force equal to ma, where a is the centripetal acceleration ($a_c = m v^2 / r$)

13. Jul 18, 2006

### garyljc

Another question
does it matter if i find the compenent parallel to the rod or parallel to the weight ?

14. Jul 18, 2006

### Staff: Mentor

I'll answer that question with a question: Which direction does the centripetal acceleration point?