# How to calculate velocity in this question

1. May 31, 2010

### Gardalay

1. The problem statement, all variables and given/known data

a pendulum of mass 1.0kg and length 1.0m is swung from a height of 0.110m above its equilibrium position. Find the tension of the pendulum string at the bottom of its swing.

Formula:
T=mv^2/r+mg

How do i get the V^2?

2. Jun 1, 2010

### Gardalay

I'm trying to find the tension...I don't have the # for T. do i just use v^2=Vi^2+2ad? and make velocity initial 0?

3. Jun 1, 2010

### pat666

no its periodic motion not projectile

4. Jun 1, 2010

### Gardalay

what is the # for T and u?

5. Jun 1, 2010

### pat666

do you mean coefficient by #????
the velocity at the bottem is given by -wAsin(wt) and from x=Acos(wt) you know that wt =1 since velocity is max at the bottem

6. Jun 1, 2010

### Gardalay

yeah...what's the coefficient for T and u

7. Jun 1, 2010

### pat666

dont worry about what i said before it was for something else,,,, I get v at the bottom as 0.3445m/s, is the T in the formula up the top period or tension just asking because ive never seen it before.

8. Jun 1, 2010

### Gardalay

it's for tension

9. Jun 1, 2010

### pat666

do you get the right answer when you sub the v i found into your equation??? answer 9.93N

10. Jun 1, 2010

### Gardalay

yeah i got that answer (I don't know if it was right though since I don't have the answer key). How did you find V again? I have no idea what '-wAsin(wt) and from x=Acos(wt)' was.

11. Jun 1, 2010

### graphene

use energy conversion to get mgh = (1/2) mv^2. v is the speed at the bottom.
and then T = mg + mv^2 / r

12. Jun 1, 2010

### Gardalay

that equation doesn't give me a velocity of 0.3445m/s.
mgh=1/2mv^2
(1kg)(9.8m/s^2)(1.1m)=1/2(1kg)(v^2)
=4.64m/s

Is it 4.64 m/s?

13. Jun 1, 2010

### pat666

t my procedure should yield exactly the same answer as using energy methods, T=2pisqrt(L/G) T=2.0 f= 0.498 w=3.13 v=wA v=0.244m/s , anyone know why that doesn't work?

14. Jun 1, 2010

### pat666

from energy methods v=1.469 but im not sure i agree with it since your acceleration (9.81) is for free fall but this is not the case for pendulums which is why our answer dont line up.

there is a component of that acceleration in the x direction, im quietly confident that you cant use energy methods to solve this now,,,, more proof there is a horizontal displacement that looses energy. in conclusion you cant use energy methods!!!!!!!!!

15. Jun 1, 2010

### Gardalay

I have no idea, can someone clarify this please?

16. Jun 1, 2010

### rl.bhat

The above equation should be

(1kg)(9.8m/s^2)(0.11m)=1/2(1kg)(v^2)

17. Jun 1, 2010

### pat666

Equations you need: note w is onega or angular velocity
X(t)=Acos(wt)
now you know that at equilibrium velocity is a maximum. so at this point A=x
cos 0 =1 therefore wt=1
T(period)=2pi*sqrt(L/g)
T=2 , f=t^-1 f=0.498Hz , w =2pif, w=3.13
v(t)=wAsin(wt) as said before, wt is 1
vmax=wa
vmax=3.13*0.11
v=0.344m/s

need more clarification??? ps as i said before you cant use energy methods

18. Jun 1, 2010

### pat666

rl.bhat do you think that that will give the correct velocity at the bottom??? ie do you think you can use energy methods for this?

19. Jun 1, 2010

### rl.bhat

In the simple pendulum, for small oscillations, the total energy remains the constant.

20. Jun 1, 2010

### pat666

The energy in the system would certainly remain constant. I dont understand how mgh=1/2mv^2 would possibly work though. the velocity that gives would be the velocity if it was dropped straight down with nothing attached. the velocity i calculated is significantly different than the one using energy and this oscillation would be called a "small oscillation" can you see anything wrong with what i did??