
#1
Apr308, 09:27 PM

P: 13

1. The problem statement, all variables and given/known data
Calculate the maximum speed of 100g pendulum mass when it has a length of 100cm and an amplitude of 50cm. 2. Relevant equations I think that Eg=mgh and Ek=0.5mv^2 are related to this problem. 3. The attempt at a solution I'm not really sure how to start this problem as I don't know how to calculate the height for Eg. However I'm pretty sure that I need to use further on gh=0.5v^2 as mass cancels in this situation. Sorry that I cannot provide a full attempt, but I just don't understand part of the problem. I just tried to solve it again and that what I got: sqrt(g/L) =sqrt(9.8/1) =3.13 v(max)=(.5)(3.13) =1.57 m/s Anything right? Thank you in advance. 



#2
Apr408, 01:29 AM

P: 339

HINT:
The speed of a simple pendulum is maximum at its center. Also .. v= A * omega *cos(omega*t + phase angle ) [S.H.M. EQUATION] At center phase angle equals zero. 



#3
Apr408, 05:07 AM

Mentor
P: 40,878

Draw a diagram! 



#4
Apr408, 05:16 AM

P: 13

Pendulum Problem
So by I found out the vertical distance from the support point is 86.6cm, however it does not make sense, if the length is 100cm wouldn't the height at rest be the same?
I can continue from here but I need explanation about the height. Thank you again 



#5
Apr408, 05:37 AM

Mentor
P: 40,878





#6
Apr408, 05:45 AM

P: 13

I'm really sorry but I'm a bit confused about the wording, if the lowest point is 86.6cm then I just subtract this from 100cm therefore the height of the mass before it is released is 13.6cm? I just can't figure it out. Can you please give me a hint or some further explanation.
Thank you for you patience. 



#7
Apr408, 06:22 AM

Mentor
P: 40,878

If you are having a hard time visualizing this, draw a diagram showing the pendulum at its highest and lowest point. Note: The highest position of the mass is 86.6cm below the support, which means it is 13.4cm above the reference point. 



#8
Apr408, 06:26 AM

P: 13

Alright! :D
So now I need to calculate Eg at h=13.6cm Ek=0 before released. But how do I calculate the speed at the bottom? 



#9
Apr408, 06:28 AM

Mentor
P: 40,878

Use conservation of energy. You already gave the correct formula.




#10
Apr408, 06:29 AM

P: 13

Therefore Ek at the bottom will equal the same as Eg before release.
Ek = 0.5mv^2 Then I need to find v? 



#11
Apr408, 07:14 AM

Mentor
P: 40,878

Exactly.
Mechanical energy is conserved, so Ek(1) + Eg(1) = Ek(2) + Eg(2). Since we measure Eg from the bottom level, Eg(2) = 0; that gives you: Eg(1) = Ek(2) mgh = 0.5mv^2 



#12
Apr408, 04:04 PM

P: 13

Thank you very much!




#13
May2308, 08:01 PM

P: 2

doc could you help me out here
I am stumped... I just sent you a report on what data I have collected this far 



#14
May2308, 08:07 PM

P: 2

nope, got it now. But another note. How do I find the Acc. due to grav.???



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