Lee S.

asked • 01/15/14

i am trying to find potential and kinetic energy of a pendulum

i found the pe the mass was .1 kg x 9.8 gravity height .155 m the question is calculate the ke of pendulum at its lowest point using the formula mgh=1/2mv2

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Tom D. answered • 01/16/14

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Two parts to this question:
 
Part 1: Steve S is absolutely correct in his OP.  This is a conservation of Energy problem.  The ONLY important quantity is delta-H (The vertical displacement from initial condition to lowest point of the mass CG.  The path of the (arc) is irrelevant.  
 
1/2mV^2=mg*delta-H
Vmax = sqrt(2g*delta-H)   (which occurs at the bottom of the arc when pendulum is vertical)
delta-H is the vertical displacement of the Center of gravity of the mass (probably center of spherical weight) at it's "low_point" to its "high_point"   Let's define these two terms
low_point: Bottom of arc (pendulum is vertical)  Use this as a ZERO reference point
high_point: Top of arc (pendulum at angle theta)  Measure only the vertical displacement.  It will roughly be L(1-cos(theta)) where theta=0 when pendulum is vertical.
L: Length of bar from pivot axis to CG of the spherical mass)
 
Part 2: Andre W has the correct eqtn above.  T~2pi*sqrt(L/g)
Notes:
 
1)This is a SMALL ANGLE APPROXIMATION formula.  In reality, the solution is a sin(theta) function, but for small angles (theta~sin(theta)).  Therefore, smaller amplitudes give more accurate results.
2)The rod is presumed massless
 
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Andre W. answered • 01/15/14

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Sorry, Δh=.155-.11=.045 m, which gives you vtheoretical=.939 m/s.
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Andre W.

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My comments no longer show up. Maybe this works.
Since Δh is so small compared to the length of the pendulum, you can use the horizontal distance .5 m to approximate the arclength. Therefore, vexperimental=4(0.5)/3.26=.613 m/s, or about 65% of the theoretical velocity. The difference is due to air resistance and other frictional losses.
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01/15/14

Steve S. answered • 01/15/14

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mgh: mass times acceleration due to gravity time height is potential energy.
1/2 mv^2: half of mass times velocity squared is kinetic energy.
 
A pendulum's energy sloshes back and forth between all potential and all kinetic.
 
So the maximum kinetic energy = maximum potential energy (although they occur at different points in a cycle).
 
max kinetic energy = (0.1 kg) * (9.8 m/s^2) * (0.155 m) joules = 0.1519 joules
 
max kinetic energy = 1/2 m(v_max)^2 = 0.1519 joules
 
So v_max = sqrt( 0.1519 * 2 / 0.1 ) ≈ 1.74299 m/s
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Lee S.

got that far so how do you find the velocity at the lowest point having trouble with that, we were doing this activity in class and time for 1 period average was 3.26 do the theoretical value for velocity and what i have do not equal out
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01/15/14

Vivian L.

Steve;
mgh=1/2mv2
Your equation...
max kinetic energy = (0.1 kg) * (9.8 m/s^2) * (0.155 m) joules = 0.1519 joules
Did you omit the 1/2, or am I missing something?  It has been a while since I minored in physics.
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01/15/14

Steve S.

Vivian,
 
maximum kinetic energy = maximum potential energy
 
So I calculate the max potential energy and set the result to max kinetic energy.
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01/15/14

Lee S.

 Got that also did i time the period wrong 3.26 seconds is what i got in class today i held the weight at .5 m from equilibrium so the distance 2 meters
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01/15/14

Steve S.

Lee, I think your geometry is wrong. If the pendulum is 2 m long and you held it 1/2 m from center line when you let it go, I calculate that the height was 0.8820 m, not 0.155 m. Something's wrong.
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01/15/14

Andre W.

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Unless your pendulum was about 2.6 meters long (almost the height of a typical lab room), you measured the period wrong. Remember the formula for the period of a simple pendulum is T=2π√(L/g), so that L=g(T/2π)²=9.8(3.26/2π)²≈2.6 m. Since your pendulum wasn't a simple pendulum (which is a model), you had to measure its length, initial height, and period separately to calculate the actual maximum velocity.
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01/15/14

Lee S.

my pendulum was 2.55 m the initial height from the floor to the bottom of the weight was  .11 m and the height at .5 m was .155 m  
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01/15/14

Andre W.

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Now you need to tell us what distance the .5 m represents. Remember the pendulum travels along a circular arc; we need the arclength to find the velocity.
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01/15/14

Lee S.

i pulled the weight of the pendulum back from equilbrium .5 m and measured the height of the weight from the floor 
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01/15/14

Andre W.

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For the theoretical maximum velocity, you should take Δh=.155-.11=.055 m in the formula for potential energy, since it is the change in height that matters. Then you get v=√(2gh)=√(2*9.8*.055)=1.04 m/s.
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01/15/14

Andre W.

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Is the .5 m the horizontal distance you pulled it or more like a diagonal/hypotenuse in a right triangle? In either case, it is not the actual arclength that the pendulum traveled.
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01/15/14

Andre W.

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In the theoretical velocity you should use the change in height, Δh=.155-.1=.055 m, since it is the change in potential energy that gives you all of the kinetic energy. This gives you v=1.04 m/s.
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01/15/14

Lee S.

Horizontal on the floor
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01/15/14

Andre W.

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Sorry, Δh=.155-.11=.045 m, so vth=.939 m/s. Since Δh is so small, you can approximate the arclength with the horizontal distance, 0.5 m. Therefore, vexp=4(0.5)/3.26=0.613 m/s., so about 65% of the theoretical value. It's so much less because of air resistance and other frictional losses.
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01/15/14

Andre W.

tutor
Δh=.155-.11=.045 m, which gives you vtheoretical=.939 m/s.
Since Δh is so small compared to the length of the pendulum, you can use the horizontal distance .5 m to approximate the arclength. Therefore, vexperimental=4(0.5)/3.26=.613 m/s, or about 65% of the theoretical velocity. The difference is due to air resistance and other frictional losses.
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01/15/14

Lee S.

the worksheet i had said to time the period which is the time it takes to swing over and back and divide by the didtance
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01/15/14

Andre W.

tutor
Δh=.155-.11=.045 m, which gives you vtheoretical=.939 m/s.
Since Δh is so small compared to the length of the pendulum, you can use the horizontal distance .5 m to approximate the arclength. Therefore, vexperimental=4(0.5)/3.26=.613 m/s, or about 65% of the theoretical velocity. The difference is due to air resistance and other frictional losses.
Report

01/15/14

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