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  • Derive the equation for rotational work.
  • Calculate rotational kinetic energy.
  • Demonstrate the Law of Conservation of Energy.

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

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

Now, we solve one of the rotational kinematics equations for αθ size 12{ ital "αθ"} {} . We start with the equation

ω 2 = ω 0 2 + 2 αθ . size 12{ω rSup { size 8{2} } =ω rSub { size 8{0} rSup { size 8{2} } } +2 ital "αθ"} {}

Next, we solve for αθ size 12{ ital "αθ"} {} :

αθ = ω 2 ω 0 2 2 . size 12{ ital "αθ"= { {ω rSup { size 8{2} } - ω rSub { size 8{0} rSup { size 8{2} } } } over {2} } } {}

Substituting this into the equation for net W size 12{W} {} and gathering terms yields

net W = 1 2 2 1 2 I ω 0 2 . size 12{"net "W= { {1} over {2} } Iω rSup { size 8{2} } - { {1} over {2} } Iω rSub { size 8{0} rSup { size 8{2} } } } {}

This equation is the work-energy theorem    for rotational motion only. As you may recall, net work changes the kinetic energy of a system. Through an analogy with translational motion, we define the term 1 2 2 size 12{ left ( { {1} over {2} } right )Iω rSup { size 8{2} } } {} to be rotational kinetic energy     KE rot size 12{ ital "KE" rSub { size 8{ ital "rot"} } } {} for an object with a moment of inertia I size 12{I} {} and an angular velocity ω size 12{ω} {} :

Questions & Answers

how to prove that Newton's law of universal gravitation F = GmM ______ R²
Kaka Reply
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Olubukola Reply
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Kaka Reply
prove that centripetal force Fc = MV² ______ r
Kaka
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mitul Reply
griffts bridge derivative
Ganesh Reply
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Timothy Reply
what is a conductor
Timothy
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Timothy
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Timothy
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dust particles contains both positive and negative charge particles
Mbutene
corona charge can verify
Stephen
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Ibrahim Reply
what is frequency
Mbionyi Reply
define precision briefly
Sujitha Reply
CT scanners do not detect details smaller than about 0.5 mm. Is this limitation due to the wavelength of x rays? Explain.
MITHRA Reply
hope this helps
what's critical angle
Mahmud Reply
The Critical Angle Derivation So the critical angle is defined as the angle of incidence that provides an angle of refraction of 90-degrees. Make particular note that the critical angle is an angle of incidence value. For the water-air boundary, the critical angle is 48.6-degrees.
dude.....next time Google it
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Chidalu
pls who can give the definition of relative density?
Temiloluwa
the ratio of the density of a substance to the density of a standard, usually water for a liquid or solid, and air for a gas.
Chidalu
What is momentum
aliyu Reply
mass ×velocity
Chidalu
it is the product of mass ×velocity of an object
Chidalu
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Sean Reply
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Wat is the relationship between Instataneous velocity
Oyinlusi Reply
Instantaneous velocity is defined as the rate of change of position for a time interval which is almost equal to zero
Astronomy
The potential in a region between x= 0 and x = 6.00 m lis V= a+ bx, where a = 10.0 V and b = -7.00 V/m. Determine (a) the potential atx=0, 3.00 m, and 6.00 m and (b) the magnitude and direction of the electric ficld at x =0, 3.00 m, and 6.00 m.
Practice Key Terms 2

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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