# 10.8 Work and power for rotational motion

 Page 1 / 7
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 $d\theta$ from A to B while under the influence of a force $\stackrel{\to }{F}$ . The external force $\stackrel{\to }{F}$ is applied to point P , whose position is $\stackrel{\to }{r}$ , 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 $\stackrel{\to }{r}$ moves in a circle of radius r , and the vector $d\stackrel{\to }{s}$ is perpendicular to $\stackrel{\to }{r}.$ 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

$\stackrel{\to }{s}=\stackrel{\to }{\theta }\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{r}.$

Thus,

$d\stackrel{\to }{s}=d\left(\stackrel{\to }{\theta }\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{r}\right)=d\stackrel{\to }{\theta }\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{r}+d\stackrel{\to }{r}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{\theta }=d\stackrel{\to }{\theta }\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{r}.$

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

$W=\int \sum \stackrel{\to }{F}·d\stackrel{\to }{s}=\int \sum \stackrel{\to }{F}·\left(d\stackrel{\to }{\theta }\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{r}\right)=\int d\stackrel{\to }{\theta }·\left(\stackrel{\to }{r}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\sum \stackrel{\to }{F}\right)$

where we used the identity $\stackrel{\to }{a}·\left(\stackrel{\to }{b}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{c}\right)=\stackrel{\to }{b}·\left(\stackrel{\to }{c}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{a}\right)$ . Noting that $\left(\stackrel{\to }{r}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\sum \stackrel{\to }{F}\right)=\sum \stackrel{\to }{\tau }$ , we arrive at the expression for the rotational work    done on a rigid body:

$W=\int \sum \stackrel{\to }{\tau }·d\stackrel{\to }{\theta }.$

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

$dW=\left(\sum _{i}{\tau }_{i}\right)d\theta$

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 $d\theta$ . The quantity $\left(\sum _{i}{\tau }_{i}\right)$ 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 ${W}_{i}=\text{Δ}{K}_{i}$ 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

${W}_{AB}={K}_{B}-{K}_{A}$

where

$K=\frac{1}{2}I{\omega }^{2}$

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

${W}_{AB}=\underset{{\theta }_{A}}{\overset{{\theta }_{B}}{\int }}\left(\sum _{i}{\tau }_{i}\right)d\theta .$

#### Questions & Answers

principle of superposition?
principle of superposition allows us to find the electric field on a charge by finding the x and y components
Kidus
Ok i got a question I'm not asking how gravity works. I would like to know why gravity works. like why is gravity the way it is. What is the true nature of gravity?
gravity pulls towards a mass...like every object is pulled towards earth
Ashok
An automobile traveling with an initial velocity of 25m/s is accelerated to 35m/s in 6s,the wheel of the automobile is 80cm in diameter. find * The angular acceleration
what is the formula for pressure?
force/area
Kidus
force is newtom
Kidus
and area is meter squared
Kidus
so in SI units pressure is N/m^2
Kidus
In customary United States units pressure is lb/in^2. pound per square inch
Kidus
who is Newton?
scientist
Jeevan
a scientist
Peter
that discovered law of motion
Peter
ok
John
but who is Isaac newton?
John
a postmodernist would say that he did not discover them, he made them up and they're not actually a reality in itself, but a mere construct by which we decided to observe the word around us
elo
how?
Qhoshe
Besides his work on universal gravitation (gravity), Newton developed the 3 laws of motion which form the basic principles of modern physics. His discovery of calculus led the way to more powerful methods of solving mathematical problems. His work in optics included the study of white light and
Daniel
and the color spectrum
Daniel
what is a scalar quantity
scalar: are quantity have numerical value
muslim
is that a better way in defining scalar quantity
Peter
thanks
muslim
quantity that has magnitude but no direction
Peter
upward force and downward force lift
upward force and downward force on lift
hi
Etini
hi
elo
hy
Xander
Hello
Jux_dob
hi
Peter
Helo
Tobi
Yes what about it?
Daniel
what's the answer? I can't get it
what is the question again?
Sallieu
What's this conversation?
Zareen
what is catenation? and give examples
sununu
How many kilometres in 1 mile
Nessy
1.609km in 1mile
Faqir
what's the si unit of impulse
The Newton second (N•s)
Ethan
what is the s. I unit of current
Amphere(A)
imam
thanks man
Roland
u r welcome
imam
the velocity of a boat related to water is 3i+4j and that of water related to earth is i-3j. what is the velocity of the boat relative to earth.If unit vector i and j represent 1km/hour east and north respectively
what is head to tail rule?
Explain Head to tail rule?
kinza
what is the guess theorem
viva question and answer on practical youngs modulus by streching
send me vvi que
rupesh
a car can cover a distance of 522km on 36 Liter's of petrol, how far can it travel on 14 liter of petrol.
Isaac
yoo the ans is 193
Joseph
whats a two dimensional force
what are two dimensional force?  By  By     By 