Simple Pendulum Acceleration

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  • #1
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Homework Statement



A simple pendulum of mass 19 kg is displaced an angle of 13 degrees from vertical and released. It now has a period of 2 sec.
a) Its mass is doubled. What is its period now?
T2m = 2 sec

b) Its length is doubled. What is its period now?
T2L = 2.828 s

c) What if the original pendulum is only displaced a distance of 6.5 before being released. What is its period now?
TΘ/2 = 2 sec

d) The original pendulum is taken to a planet where g = 15 m/s2. What is its period on that planet?
T = 1.6174 s

e) Let's go back to the original pendulum of mass 19 kg with a period of 2 sec, displaced an angle of 13 degrees from the vertical. What would its acceleration be in the vertical (y) direction as it reachs the lowest point on its swing?
ay = _______ m/sec2

Homework Equations



T=2[tex]\Pi[/tex][tex]\sqrt{L/g}[/tex]

The Attempt at a Solution



I solved all of the questions except for part e.
I thought it might be zero, but that didn't work.
I'm not sure what to do since I don't know the length or anything.

Thanks for your help in advance!
 

Answers and Replies

  • #2
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The bob is in circular motion, right? So how is the component of the acceleration towards the center in circular motion called? It has its own specific name. I hope you remember it :)
 
  • #3
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The bob is in circular motion, right? So how is the component of the acceleration towards the center in circular motion called? It has its own specific name. I hope you remember it :)
Centripetal acceleration? But how do I find that if I don't know the radius?
 
  • #4
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Is the radius the length of the rope? :)
 
  • #5
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Oh I see! Forgot I could solve for that!

Okay so I have the length. I solved for the torque (hopefully correctly) by doing the weight of the mass times that length. For the moment of inertia, would it just be for a point mass MR^2? When I solve it this way, my acceleration just comes out as gravity and it's incorrect.


Thank you thank you again!!
 
  • #6
798
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Okay so I have the length. I solved for the torque (hopefully correctly) by doing the weight of the mass times that length. For the moment of inertia, would it just be for a point mass MR^2? When I solve it this way, my acceleration just comes out as gravity and it's incorrect.
Oh, the torque is wrongly calculated. You must take into account the angle between the rope and the vertical direction. The torque you got is the torque at the time the rope is in horizontal position. Remember how to calculate torque? :confused:
But what you got is NOT the centripetal acceleration. It's the component perpendicular to the rope. And, again, at the time when the bob is at the horizontal position, isn't that component equal to gravitational acceleration? It's a big problem, be careful! :cry:
Luckily you found it's incorrect :approve: Cheers!
 
  • #7
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Thanks again for your help. I just can't figure this out!
 

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