# Pendulum help

386221

## Homework Statement

You are given a pendulum composed of a 0.030 kg mass on the end of a 0.60 m long massless string. If the pendulum is moved 30° from the vertical and given an initial speed of 0.39 m/s tangent to the support string and away from the vertical, how much higher relative to the release point of the pendulum swing will the pendulum rise?

KE=1/2mv^2
PE=mgL(1-cosθ)

## The Attempt at a Solution

Here is what I did. I found the velocity at the bottom of the swing.
1/2vf^2 = 1/2vi^2 + mgL(1-cosθ)
and I got vf= 1.3 m/s
Then I did the same thing again, except I set vi to 1.3 and vf = 0
1/2vi^2 = mgL(1-cosθ)
and find θ, which I got 31.1 then I subtracted Lcos30 by Lcos31.1 and got .005 m. The answer is incorrect, can anyone help? BTW I already cancelled out the mass in all the above work.

Homework Helper
Gold Member
Dearly Missed
It seems you forgot the m's in your kinetic energies??

Tanya Sharma
Hi 386221

Welcome to PF!!!

Your work looks good to me .What is the correct answer ?

Homework Helper
Gold Member
Dearly Missed
I haven't checked the data, Tanya Sharma, but it seems that OP is using 1/2v^2, rather than 1/2mv^2.
I agree that OP, apart from that seems to have chosen a correct, if somewhat overcomplicated solution since there is no need to use the intermediate position at the bottom.

Enigman

## Homework Statement

You are given a pendulum composed of a 0.030 kg mass on the end of a 0.60 m long massless string. If the pendulum is moved 30° from the vertical and given an initial speed of 0.39 m/s tangent to the support string and away from the vertical, how much higher relative to the release point of the pendulum swing will the pendulum rise?

KE=1/2mv^2
PE=mgL(1-cosθ)

## The Attempt at a Solution

Here is what I did. I found the velocity at the bottom of the swing.
1/2vf^2 = 1/2vi^2 + mgL(1-cosθ)
and I got vf= 1.3 m/s
Then I did the same thing again, except I set vi to 1.3 and vf = 0
1/2vi^2 = mgL(1-cosθ)
and find θ, which I got 31.1 then I subtracted Lcos30 by Lcos31.1 and got .005 m. The answer is incorrect, can anyone help? BTW I already cancelled out the mass in all the above work.
check again.

Tanya Sharma
I haven't checked the data, Tanya Sharma, but it seems that OP is using 1/2v^2, rather than 1/2mv^2.
I agree that OP, apart from that seems to have chosen a correct, if somewhat overcomplicated solution since there is no need to use the intermediate position at the bottom.

I checked the data before replying.OP has mentioned that he cancelled the masses in his work,which he/she did.He/She has erroneously written 1/2v^2 instead of 1/2mv^2.

Homework Helper
Gold Member
Dearly Missed
`Well, then his "incorrect" answer might just mean that he rounded off numbers in an incorrect way in his intermediate step.
Thank you, Tanya Sharma, for having checked the data!

Enigman
For an easier method just set potential energy at the point of beginning of motion as 0.
##U_i+K_i=U_f+K_f##
##K_i=U_f##
##1/2(mv^2)=mg\Delta h##

Homework Helper
Gold Member
Dearly Missed
There is a couple of sources for errors which might explain the incorrectness of OP's answer:
1. He might at some point have made a mix-up of radians versus degrees.
2. Or, and I think this is the likeliest source for the error error: the arccosine function is typically defined relative to the interval 0 to pi, while in this problem, the answer might have be given for angles in the interval -pi/2 to pi/2 instead. That is, a minus sign might be missing in OP's solution for the angle.

Enigman
The angle is slightly off...31.0323136133
(and there was another error in cos31.1)

386221
I tried changing the second angle to 31.03 and got delta h of .009 and it was wrong. I also tried using 1/2mv^2=mgh and got .0077 for h and that was wrong as well. I honestly have no clue on how to proceed all my work seems correct...