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Energy Transformation for a Pendulum

A GIF Animation

The motion of a pendulum is a classic example of mechanical energy conservation. A pendulum consists of a mass (known as a bob) attached by a string to a pivot point. As the pendulum moves it sweeps out a circular arc, moving back and forth in a periodic fashion. Neglecting air resistance (which would indeed be small for an aerodynamically shaped bob), there are only two forces acting upon the pendulum bob. One force is gravity; the force of gravity acts in a downward direction and does work upon the pendulum bob. However, gravity is an internal force (or conservative force) and thus does not serve to change the total amount of mechanical energy of the bob. The other force acting upon the bob is the force of tension. Tension is an external force and if it did do work upon the pendulum bob it would indeed serve to change the total mechanical energy of the bob. The force of tension does not do work since it always acts in a direction perpendicular to the motion of the bob. At all points in the trajectory of the pendulum bob, the angle between the force of tension and its instantaneous displacement is 90 degrees. Thus, the force of tension does not do work upon the bob [F*d*cosine(90 degrees) is 0 Joules].

Since there are no external forces doing work, the total mechanical energy of the pendulum bob is conserved. The conservation of mechanical energy is demonstrated in the animation below. Observe the KE and PE bars of the bar chart; their sum is a constant value.

Observe that the falling motion of the bob is accompanied by an increase in speed. As the bob loses height and PE, it gains speed and KE; yet the total of the two forms of mechanical energy is conserved.

To test you understanding of the conservation principle, use the heights and the speeds given in the table below to fill in the remaining cells at the various locations in a 0.200-kg bob's trajectory. Use the pop-up menus to check your work.

(Note: mass = 0.200 kg)

h (m)	Speed (m/s)	PE (J)	KE (J)	TME (J)
2.000	0.0 m/s	Answer PE = m*g*h g=9.8 m/s/s m=0.200 kg h=2.000 m so PE=3.920 J	Answer KE=.5*m*v^2 m=0.2 and v=0.0 so KE = 0 J	Answer TME=PE+KE PE=3.920 J KE=0 J so TME=3.920 J
1.490	Answer First find PE, then KE, then speed. Since KE=.5*m*v^2, v=SQRT(2*KE/m). With KE=1.00 J m=0.200 kg Find v = 3.162 m/s.	Answer PE = m*g*h g=9.8 m/s/s m=0.200 kg h=1.490 m so PE=2.920 J	Answer TME=PE+KE So KE = TME - PE TME = 3.920 J PE=2.920 J so KE = 1.000 J	3.920 J
1.128	Answer Find PE, KE, then speed. KE=.5*m*v^2, so v=SQRT(2*KE/m). KE=1.709 J and m=0.200 kg So v = 4.134 m/s.	Answer PE = m*g*h g=9.8 m/s/s m=0.200 kg h=1.128 m so PE=2.211 J	Answer TME=PE+KE So KE = TME - PE TME = 3.920 J PE=2.211 J so KE=1.709 J	Answer TME is conserved. It will be the same as it was initially: 3.920 J.
0.897	Answer Find PE, KE, then speed. KE=.5*m*v^2, so v=SQRT(2*KE/m). KE=2.172 J and m=0.200 kg, so v = 4.660 m/s.	Answer PE = m*g*h g=9.8 m/s/s m=0.200 kg h=0.897 m so PE=1.748 J	Answer TME=PE+KE So KE = TME - PE TME = 3.920 J PE=1.748 J so KE=2.172 J	Answer TME is conserved. It will be the same as it was initially: 3.920 J.
0.357	Answer Find PE, KE, then speed. KE=.5*m*v^2, so v=SQRT(2*KE/m). KE=3.220 J and m=0.200 kg, so v = 5.675 m/s.	Answer PE = m*g*h g=9.8 m/s/s m=0.200 kg h=0.357 m so PE=0.700 J	Answer TME=PE+KE So KE = TME - PE TME = 3.920 J PE=0.700 J so KE=3.220 J	Answer TME is conserved. It will be the same as it was initially: 3.920 J.

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For more information on physical descriptions of motion, visit The Physics Classroom. Specific information is available there on the following topics:

• Potential Energy
• Kinetic Energy
• Mechanical Energy
• Internal vs. External Forces
• Work
• Work-Energy Theorem
• Work-Energy Bar Charts

Other animations can be seen at the Multimedia Physics Studios. Other useful resources regarding the physics of motion is available through the Glenbrook South Physics Home Page.

This page was created by Tom Henderson of Glenbrook South High School.

Comments and suggestions can be sent by e-mail to Tom Henderson.

A hearty thanks is due to lab assistant Carl Bobis for his

assistance with the graphics and GIF animation.

This page last updated on 4/4/98.