Calculating the Acceleration of a Simple Pendulum

AI Thread Summary
The discussion focuses on calculating the speed and acceleration of a simple pendulum with a length of 4.9 m, pulled 0.5 m to one side before release. The speed at the lowest point is derived using the equation for maximum velocity, resulting in approximately 9.287 m/s. The acceleration at this point is zero, as maximum acceleration occurs at the amplitude of the swing. The work-energy theorem is mentioned as a method to relate potential and kinetic energy in this context. The conversation emphasizes the principles of Simple Harmonic Motion (SHM) and the assumptions involved in pendulum motion.
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Homework Statement



A simple pendulum has a length of 4.9 m and is pulled a distance y = 0.5 m to one side and then released.

(a) What is the speed of the pendulum when it passes through the lowest point on its trajectory?

(b) What is its acceleration when it passes through the lowest point on its trajectory?


Homework Equations



I think this is the equation I am supposed to use? omega=sqrt(2(9.8)(L(1-cos(a))) where omega is velocity max and L is length of pendulum, a is angle between Pendulum and its starting position

Not totally sure what to do with equations...

The Attempt at a Solution



I tried solving for a and I got 1.262896 radians
then a final speed of 9.287 m/s
 
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corey2014 said:

Homework Statement



A simple pendulum has a length of 4.9 m and is pulled a distance y = 0.5 m to one side and then released.

(a) What is the speed of the pendulum when it passes through the lowest point on its trajectory?

(b) What is its acceleration when it passes through the lowest point on its trajectory?


Homework Equations



I think this is the equation I am supposed to use? omega=sqrt(2(9.8)(L(1-cos(a))) where omega is velocity max and L is length of pendulum, a is angle between Pendulum and its starting position

Not totally sure what to do with equations...

The Attempt at a Solution



I tried solving for a and I got 1.262896 radians
then a final speed of 9.287 m/s

Hopefully the question had a diagram with it, which will enable you to answer the following question: Was the 0.5 m the length of the arc along which the bob was drawn back, or is it the horizontal distance to the side, that the bob is drawn?
 
Use the work energy theorem.
 
There is no diagram but because it says y= can't I assume that its in the Y direction? And what is that theorem?
 
I think it means the horizontal distance it is displaced, the work energy theorem says that the total change in kinetic energy of an object is equal to the sum of the work done on that object. In this case the work done by gravity in going from the highest position to the lowest is equal to the kinetic energy at the lowest (since its speed was 0 in the highest position).
 
This is a problem about Simple Harmonic Motion (SHM)
A simple pendulum can be assumed to perform SHM.
The amplitude, A, of the motion is 0.5m.
The Time period and therefore the angular velocity can be found from the pendulum equation
T =2∏√l/g, ω=2∏/T
In SHM max velocity occurs when amplitude = 0 and is given by ωA
Acceleration when amplitude = 0 is ZERO
Max acceleration = ω^2 A and occurs when displacement = A

All of this makes the usual assumptions about a simple pendulum... small angle etc.
 
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