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By the end of this section, you will be able to:
  • Use the work-energy theorem to analyze rotation to find the work done on a system when it is rotated about a fixed axis for a finite angular displacement
  • Solve for the angular velocity of a rotating rigid body using the work-energy theorem
  • Find the power delivered to a rotating rigid body given the applied torque and angular velocity
  • Summarize the rotational variables and equations and relate them to their translational counterparts

Thus far in the chapter, we have extensively addressed kinematics and dynamics for rotating rigid bodies around a fixed axis. In this final section, we define work and power within the context of rotation about a fixed axis, which has applications to both physics and engineering. The discussion of work and power makes our treatment of rotational motion almost complete, with the exception of rolling motion and angular momentum, which are discussed in Angular Momentum . We begin this section with a treatment of the work-energy theorem for rotation.

Work for rotational motion

Now that we have determined how to calculate kinetic energy for rotating rigid bodies, we can proceed with a discussion of the work done on a rigid body rotating about a fixed axis. [link] shows a rigid body that has rotated through an angle d θ from A to B while under the influence of a force F . The external force F is applied to point P , whose position is r , and the rigid body is constrained to rotate about a fixed axis that is perpendicular to the page and passes through O . The rotational axis is fixed, so the vector r moves in a circle of radius r , and the vector d s is perpendicular to r .

Figure shows the rigid body is constrained to rotate about a fixed axis that is perpendicular to the page and passes through a point labeled as O. The rotational axis is fixed, so the vector r moves in a circle of radius r, and the vector ds is perpendicular to vector r. An external force F is applied to point P and makes rigid body rotates through an angle dtheta.
A rigid body rotates through an angle d θ from A to B by the action of an external force F applied to point P .

From [link] , we have

s = θ × r .

Thus,

d s = d ( θ × r ) = d θ × r + d r × θ = d θ × r .

Note that d r is zero because r is fixed on the rigid body from the origin O to point P . Using the definition of work, we obtain

W = F · d s = F · ( d θ × r ) = d θ · ( r × F )

where we used the identity a · ( b × c ) = b · ( c × a ) . Noting that ( r × F ) = τ , we arrive at the expression for the rotational work    done on a rigid body:

W = τ · d θ .

The total work done on a rigid body is the sum of the torques integrated over the angle through which the body rotates . The incremental work is

d W = ( i τ i ) d θ

where we have taken the dot product in [link] , leaving only torques along the axis of rotation. In a rigid body, all particles rotate through the same angle; thus the work of every external force is equal to the torque times the common incremental angle d θ . The quantity ( i τ i ) is the net torque on the body due to external forces.

Similarly, we found the kinetic energy of a rigid body rotating around a fixed axis by summing the kinetic energy of each particle that makes up the rigid body. Since the work-energy theorem W i = Δ K i is valid for each particle, it is valid for the sum of the particles and the entire body.

Work-energy theorem for rotation

The work-energy theorem for a rigid body rotating around a fixed axis is

W A B = K B K A

where

K = 1 2 I ω 2

and the rotational work done by a net force rotating a body from point A to point B is

W A B = θ A θ B ( i τ i ) d θ .

Questions & Answers

What is a volt equal to?
Clifton Reply
list and explain the 3 ways of charging a conductor
Chidimma Reply
conduction convention rubbing
Asdesaw
formula of magnetic field
Yonas Reply
Integral of a vector
Rahat Reply
define surface integral of a vector?
Rahat
the number of degree freedom of a rigid body in2-dimantion is:
Raj Reply
1
Nathan
A block (A) of weight 5 kN is to be raised by means of a 20° wedge (B) by the application of a horizontal force (P) as shown in Fig.1. The block A is constrained to move vertically by the application of a horizontal force (S). Find the magnitude of the forces F and S, when the coefficient of fricti
Danilo
A body receives impulses of 24Ns and 35Ns inclined 55 degree to each other. calculate the total impulse
Sukpen Reply
A body receives impulses of 24Ns and 35Ns inclined 55 degree to each other. calculate the total impulse
Previous
twenty four square plus thirty-five square minus to multiple thirty five twenty four and equal answer number square Via this equation defined Total Total impulse
Cemal
why simple pendulum do not vibrate indefinitely?
Zirmal
define integral vector
Rahat
what is matar
Abdulrahman Reply
define surface integral vector?
Rahat
The uniform boom shown below weighs 500 N, and the object hanging from its right end weighs 400 N. The boom is supported by a light cable and by a hinge at the wall. Calculate the tension in the cable and the force on the hinge on the boom. Does the force on the hinge act along the boom?
Jave Reply
A 11.0-m boom, AB , of a crane lifting a 3000-kg load is shown below. The center of mass of the boom is at its geometric center, and the mass of the boom is 800 kg. For the position shown, calculate tension T in the cable and the force at the axle A .
Jave
what is the S.I unit of coefficient of viscosity
Biam Reply
Derived the formula of Newton's law of universal gravitation Fg=G(M1M2)/R2
Monychol Reply
hi
Asdesaw
yes
Cemal
a non-uniform boom of a crane 15m long, weighs 2800nts, with its center of gravity at 40% of its lenght from the hingr support. the boom is attached to a hinge at the lower end. rhe boom, which mAKES A 60% ANGLE WITH THE HORIZONTAL IS SUPPORTED BY A HORIZONTAL GUY WIRE AT ITS UPPER END. IF A LOAD OF 5000Nts is hung at the upper end of the boom, find the tension in the guywire and the components of the reaction at the hinge.
dangly Reply
what is the centripetal force
Malok Reply
Of?
John
centripetal force of attraction that pulls a body that is traversing round the orbit of a circle toward the center of the circle. Fc = MV²/r
Sampson
centripetal force is the force of attraction that pulls a body that is traversing round the orbit of a circle toward the center of the circle. Fc = MV²/r
Sampson
I do believe the formula for centripetal force is F=MA or F=m(v^2/r)
John
I mean the formula is Fc= Mass multiplied by square of velocity all over the Radius of the circle
Sampson
Yes
John
The force is equal to the mass times the velocity squared divided by the radius
John
That's the current chapter I'm on in my engineering physics class
John
Centripetal force is a force of attraction which keeps an object round the orbit towards the center of a circle. Mathematically Fc=mv²/r
Adebileje
In Example, we calculated the final speed of a roller coaster that descended 20 m in height and had an initial speed of 5 m/s downhill. Suppose the roller coaster had had an initial speed of 5 m/s uphill instead, and it coasted uphill, stopped, and then rolled back down to a final point 20 m bel
tan Reply
A steel lift column in a service station is 4 meter long and .2 meter in diameter. Young's modulus for steel is 20 X 1010N/m2.  By how much does the column shrink when a 5000- kg truck is on it?
Andiswa Reply
hi
Abdulrahman
mola mass
Abdulrahman
hi
Asdesaw
what exactly is a transverse wave
Dharmee Reply
Practice Key Terms 2

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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