High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize AP Physics C: Mechanics Textbook > Unit 5: Rotation (14-20%)

Rotational Work and Torque

Rotational Kinetic EnergyPrevious Section
Angular MomentumNext Section
Popular Courses
Find My Course
0:00 / 0:00

Rotational Work and Torque


When we apply a torque to an object, we are performing work. The object is moving in angular variables rather than translational. So the work done by a torque caused by a force which is along the angular displacement is:

W =τ(θf−θi)\boxed{W\ =\tau (\theta_f-\theta_i)}
Written as an integral:
W=∫τ⃗.dθ⃗\boxed{W = \int{\vec{\tau}.d\vec{\theta}}}













Wize Tip
If the force is opposite the direction of angular displacement, then the torque is doing negative work. If they are in the same direction, then positive work.


Work-Energy Theorem for Rotational Motion

Work-kinetic energy theorem for rotational motion:
W=12Iωf2−12Iωi2\boxed{W=\frac{1}{2}I\omega_f^2-\frac{1}{2}I\omega_i^2}

Watch Out!
"Rolling" motion implies frictional force does no work on the rolling object. This leads to Mechanical Energy being conserved


0:00 / 0:00

Example: Rotating Wheel Kinetic Energy


A wheel whose rotational inertia is 10 kg.m2 starts from rest and accelerates under a constant torque of 3.0 N.m for 8.0 seconds. What is the wheel's rotational kinetic energy at the end of the 8 seconds?

Solution:
Here are the information given to us:
I=5kgm2  ωi=0τ=3Nm I = 5kgm^2 \ \ \newline \omega_i=0 \newline \tau = 3Nm
Since the wheel starts at rest, its initial kinetic energy is zero. So, we can use work-energy theorem to find change in the kinetic energy of the wheel which is its final rotational kinetic energy in this case.

W=Kf−KiW=K_f-K_i
We can find work by knowing the torque acting on the wheel using:
W=τΔθW=\tau \Delta\theta
So, we need to first find Δθ\Delta\thetausing rotational kinematic equations.

τ=Iα\tau = I\alpha
α=τI=310  rad/s2\alpha = \dfrac{\tau}{I} =\dfrac{3}{10} \ \ rad/s^2
Δθ=12αt2+ωit=1231082+0=9.6  rad\Delta\theta=\frac{1}{2}\alpha t^2+\omega_it=\frac{1}{2}\frac{3}{10}8^2+0=9.6 \ \ rad

Thus: W=Kf−0=3(9.6)=28.8  jW=K_f-0=3 (9.6)=28.8 \ \ j

So, the rotational kinetic energy after 8s is 28.8 j28.8 \ j.
A thin rod with mass of M and length of L is hinged to the wall at one end. The rod is held horizontally and then released. What is the speed of the tip of the rod as it hits the wall? Assume the hinge is frictionless. The axis of rotation for a thin rod about one of its ends is 13ML2\frac{1}{3}ML^2.

Hint: You should take the CM into consideration for problems of this kind.
This is a classic problem - professors love to ask questions about it!

Consider any round thing you can think of rolling down a hill. Here, we will use four objects: a ring (I=MR2I=MR^2), a disk (I=12MR2I=\frac12MR^2), a solid sphere (I=25MR2I=\frac25MR^2), and a hollow sphere (I=23MR2I=\frac23MR^2).

You release them from rest at the same time from height h and they all roll without slipping (that is, there is enough static friction to allow for this). The angle of the incline is θ\theta.

The static friction for this case is f=mgsin⁡θ1+mR2If=\frac{mg\sin\theta}{1+\frac{mR^2}{I}}.






a) Which one gets to the bottom of the slope first? In what order do the four objects finish?