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Learning objectives

By the end of this section, you will be able to:

  • Derive the equation for rotational work.
  • Calculate rotational kinetic energy.
  • Demonstrate the law of conservation of energy.

The information presented in this section supports the following AP® learning objectives and science practices:

  • 3.F.2.1 The student is able to make predictions about the change in the angular velocity about an axis for an object when forces exerted on the object cause a torque about that axis. (S.P. 6.4)
  • 3.F.2.2 The student is able to plan data collection and analysis strategies designed to test the relationship between a torque exerted on an object and the change in angular velocity of that object about an axis. (S.P. 4.1, 4.2, 5.1)

In this module, we will learn about work and energy associated with rotational motion. [link] shows a worker using an electric grindstone propelled by a motor. Sparks are flying, and noise and vibration are created as layers of steel are pared from the pole. The stone continues to turn even after the motor is turned off, but it is eventually brought to a stop by friction. Clearly, the motor had to work to get the stone spinning. This work went into heat, light, sound, vibration, and considerable rotational kinetic energy    .

The figure shows a mechanic cutting metal with a metal grinder. The sparks are emerging from the point of contact and jumping off tangentially from the cutter.
The motor works in spinning the grindstone, giving it rotational kinetic energy. That energy is then converted to heat, light, sound, and vibration. (credit: U.S. Navy photo by Mass Communication Specialist Seaman Zachary David Bell)

Work must be done to rotate objects such as grindstones or merry-go-rounds. Work was defined in Uniform Circular Motion and Gravitation for translational motion, and we can build on that knowledge when considering work done in rotational motion. The simplest rotational situation is one in which the net force is exerted perpendicular to the radius of a disk (as shown in [link] ) and remains perpendicular as the disk starts to rotate. The force is parallel to the displacement, and so the net work done is the product of the force times the arc length traveled:

net W = ( net F ) Δ s . size 12{"net "W= left ("net "F right ) cdot Δs} {}

To get torque and other rotational quantities into the equation, we multiply and divide the right-hand side of the equation by r size 12{r} {} , and gather terms:

net W = ( r net F ) Δ s r . size 12{"net"W= left (r" net "F right ) { {Δs} over {r} } } {}

We recognize that r net F = net τ size 12{r" net "F=" net "τ} {} and Δ s / r = θ size 12{Δs/r=θ} {} , so that

net W = net τ θ . size 12{"net "W= left ("net "τ right )θ} {}

This equation is the expression for rotational work. It is very similar to the familiar definition of translational work as force multiplied by distance. Here, torque is analogous to force, and angle is analogous to distance. The equation net W = net τ θ size 12{"net "W= left ("net "τ right )θ} {} is valid in general, even though it was derived for a special case.

To get an expression for rotational kinetic energy, we must again perform some algebraic manipulations. The first step is to note that net τ = size 12{"net "W=Iα} {} , so that

net W = I αθ . size 12{"net "W=I ital "αθ"} {}
The figure shows a circular disc of radius r. A net force F is applied perpendicular to the radius, rotating the disc in an anti-clockwise direction and producing a displacement equal to delta S, in a direction parallel to the direction of the force applied. The angle covered is theta.
The net force on this disk is kept perpendicular to its radius as the force causes the disk to rotate. The net work done is thus net F Δ s size 12{ left ("net "F right ) cdot Δs} {} . The net work goes into rotational kinetic energy.

Making connections

qWork and energy in rotational motion are completely analogous to work and energy in translational motion, first presented in Uniform Circular Motion and Gravitation .

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
tijani
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John Reply
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Siyaka Reply
A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
Jude Reply
Can you compute that for me. Ty
Jude
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David Reply
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David
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emma Reply
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what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
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Adjanou
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Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
Sahid Reply
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
Ryan
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Maurice Reply
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Maurice
answer
Magreth
progressive wave
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Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
yasuo Reply
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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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