Simple Pendulum Acceleration: Solving for Period and Mass Changes

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SUMMARY

The discussion focuses on the calculations related to a simple pendulum with a mass of 19 kg and an initial period of 2 seconds. Key findings include that doubling the mass does not change the period, while doubling the length increases the period to approximately 2.828 seconds. The period on a planet with a gravitational acceleration of 15 m/s² is calculated to be 1.6174 seconds. The user struggles with calculating vertical acceleration at the lowest point of the swing, mistakenly equating it to gravitational acceleration and needing clarification on torque and centripetal acceleration.

PREREQUISITES
  • Understanding of simple harmonic motion and pendulum mechanics
  • Familiarity with the formula T=2π√(L/g) for pendulum period calculation
  • Knowledge of centripetal acceleration and its relation to circular motion
  • Basic principles of torque and moment of inertia for point masses
NEXT STEPS
  • Study the derivation and implications of the pendulum period formula T=2π√(L/g)
  • Learn about centripetal acceleration and its calculation in circular motion contexts
  • Research torque calculations, particularly in relation to angles and forces
  • Explore the effects of varying gravitational acceleration on pendulum motion
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators seeking to clarify pendulum dynamics and related calculations.

SuperCass
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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\Pi\sqrt{L/g}

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!
 
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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 :)
 
hikaru1221 said:
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?
 
Is the radius the length of the rope? :)
 
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!
 
SuperCass said:
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!
 
Thanks again for your help. I just can't figure this out!
 

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