Pendulum Rise from Release Point: 0.005 m

[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?

Homework Equations

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 canceled out the mass in all the above work.

 

[arildno]

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 ?

 

[arildno]

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?

Homework Equations

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 canceled 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 canceled the masses in his work,which he/she did.He/She has erroneously written 1/2v^2 instead of 1/2mv^2.

 

[arildno]

`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.
$U_i+K_i=U_f+K_f$
$K_i=U_f$
$1/2(mv^2)=mg\Delta h$

 

[arildno]

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...