If we push on an object in the forward direction while the object is
moving forward, we do positive work on the object. The object accelerates,
because we are pushing on it. F = ma. It gains kinetic energy. The
**translational kinetic energy** of an
object with mass m, whose center of mass is moving with speed v is K = ½mv^{2}.

Translational kinetic energy = ½ mass * speed^{2}

Kinetic energy increases quadratically with speed. When the speed of a car doubles, its energy increases by a factor of four.

A rotating object also has kinetic energy. When an object is rotating
about its center of mass, its **rotational kinetic
energy **is K = ½Iω^{2}.

Rotational kinetic energy = ½ moment of inertia * (angular speed)^{2}.

When the angular velocity of a spinning wheel doubles, its kinetic energy
increases by a factor of four.

When an object has translational as well as rotational motion, we can look at
the motion of the center of mass and the motion about the center of mass
separately. The total kinetic energy is the sum of the translational kinetic energy of the center
of mass (CM) and the rotational kinetic energy about
the CM.

Consider a wheel of radius r and mass m rolling on a flat surface in the
x-direction.

The displacement Δx and the angular displacement Δθ are related through
Δx = rΔθ.

The magnitudes of the linear velocity and the angular velocity are related through v_{CM }= rω.

The kinetic energy of the disk is the sum of the kinetic energy of the motion of the
center of mass ½mv_{CM}^{2 }= ½mr^{2}ω^{2},
and the kinetic energy of the motion about the center of mass, ½Iω^{2}.

Thee total kinetic energy is

KE_{tot} = ½mr^{2}ω^{2}
+ ½Iω^{2} = ½[mr^{2} + I]ω^{2} = ½[m + I/r^{2}]v^{2}.

Suppose you are designing a race bicycle and it comes time to work on the wheels. You are told that the wheels need to be of a certain mass but you may design them either as wheels with spokes (like traditional bike wheels) or you may make them as having solid rims all the way through. Which design would you pick given that the racing aspect of the machine is the most important? Please explain!

By clicking the button below, you
can play or to step through a video clip frame-by-frame. Each step
corresponds to a time interval of (1/30) s. In the clip the same torque acts on objects with
different moments of inertia. The torque is the product of a weight and a
small lever arm. The moment of inertia of the ruler-like object changes
because masses are added at larger distances away from the center.
When the weight has dropped through the same distance Δy,
the same work has been done and the system has the same kinetic energy, since it
starts from rest. Neglecting friction W = τΔθ = mgΔy = ½Iω^{2}.
However, the system with the larger moment of inertia I has the smaller angular
speed ω. (Compare the angles through which the ruler turns per step without
attached masses and with masses attached at different locations.)

Three particles are connected by rigid rods of negligible mass lying along
the y-axis as shown.

If the system rotates about the x-axis with angular speed of 2 rad/s,
find

(a) the moment of inertia about the x-axis and the total rotational
kinetic energy evaluated from ½Iω^{2},
and

(b) the linear speed of each particle and the total kinetic energy evaluated
from Σ½m_{i}v_{i}^{2}.

Solution:

(a) The moment of inertia is
I = ∑m_{i}r_{i}^{2}.
Here r_{i} is the perpendicular distance of particle i from
the x-axis.

I = (4 kg)(9 m^{2})^{ }+
(2 kg)(4 m^{2})^{ }+
(3 kg)(16 m^{2})^{ }= 92 kgm^{2}.

The rotational kinetic energy is K = ½Iω^{2
}= 46*4/s^{2 }= 184 J.

(b) The linear speed of particle i is v_{i
}= ωr_{i}.

The linear speed of the 4 kg mass is v = 6 m/s, and its kinetic energy is ½mv^{2
}= 72 J.

The linear speed of the 2 kg mass is v = 4 m/s, and its kinetic energy is ½mv^{2
}= 16 J.

The linear speed of the 3 kg mass is v = 8 m/s, and its kinetic energy is ½mv^{2
}= 96 J.

The sum of the kinetic energies of the three particles is 184 J.

**Angular momentum** about an axis is a measure of an objects rotational motion
about this axis. For rotations about a symmetry axis of an object, the angular momentum **L** is defined as the product of an
object's moment of inertia I times its angular velocity **ω** about the chosen axis.

**L** = I**ω**.

A light rod 1 m in length rotates in the xy plane about a pivot through the
rod's center. Two particles of mass 4 kg and 3 kg are connected to its
ends. Determine the angular momentum of the system at the instant the
speed of each particle is 5 m/s.

Solution:

We assume that the
mass and moment of inertia of the rod can be neglected.

Let the z-axis pass through the center of the rod and point out of the page.

The moment of inertia of the system about the z-axis is

3 kg*(0.5 m)^{2 }+ 4 kg*(0.5 m)^{2
}= 1.75 kgm^{2}.

The angular velocity of the system is
**ω **= (v/r)**k **= ((5 m/s)/(0.5 m))**k **= (10/s)**k**.

Here
**k** is a unit vector or direction indicator pointing in the
z-direction (out of the page).

The angular momentum of the system is
**L **= I**ω **= (17.5 kgm^{2}/s)**k**.

Angular momentum is a vector. For a single particle its direction is the direction of the angular
velocity (given by the right hand rule). The angular momentum of an object is
changed by giving it an **angular impulse**. An
angular impulse Δ**L** is a change in angular
momentum. You give an object an angular impulse by letting a torque act on it
for a time interval Δt.

Δ**L** = **τ**Δt

angular impulse = torque * time

If an object has many independently rotating parts, the total angular momentum of the object is the sum of the angular momenta of all its parts.

You usually give your closet door a gentle push and it swings closed gently
in 5 seconds. But today you are in a rush and exert 3 times the normal
torque on it.

(a) If you push on it for the usual time with this increased torque, how will
its angular momentum differ from the usual value?

(b) How long will it take the closet door to swing closed after your hurried
push?

Solution:

(a) When you push on the door, you exert a torque (force times lever arm)
for a certain amount of time, and the door gains angular momentum
**L **= I**ω**. If you increase the force by a factor
of 3 and push for the same amount of time, the angular momentum increases by
a factor of three. The moment of inertia of the door does not change, so
its angular velocity must increase by a factor of 3.

(b) angular velocity = angular displacement / time, time = angular
displacement / angular velocity

The time it will take the door to close will decrease by a factor of three.

time = (5 s)/3 = 1.66 s .

The total angular momentum of a a single object is constant if no external torque acts on the object. An object cannot exert a torque on itself. The total angular momentum of two interacting objects is also constant if no external torque acts on the objects. Newton's third law tells us the forces the objects exert on each other are equal in magnitude and opposite in direction. The interaction forces produce torques equal in magnitude and opposite in direction. These torques change the angular momentum of each object by the same amount, but the changes will have opposite directions. When we sum them up to find the change in the total angular momentum, we obtain zero.

If no external torque acts on a system of interacting objects, then their total angular momentum is constant.

In the video clip shown below the total angular momentum of the system points upward. The person is stopping a spinning wheel and the stool starts to spin.

Some of the first few frames

As the person applies a torque to the wheel, the wheel applies a torque to the person. The magnitudes of the angular momenta of the wheel and of the person change at the same rate, but their sum remains constant.

A 60 kg woman stands at the rim of a horizontal turntable having a moment
of inertia of 500 kgm^{2}, and a radius of 2 m. The turntable is
initially at rest and is free to rotate about a frictionless vertical axis
through its center. The woman then starts walking around the rim
clockwise (as viewed from above the system) at a constant speed of 1.5 m/s
relative to the Earth.

(a) In what direction and with what angular speed does the turntable rotate?

(b) How much work does the women do to set herself and the turntable in motion?

Solution:

The system consists of the woman and the turntable. No external torques act on the system,
so the total angular momentum of the system is conserved. It is zero before the woman
starts to walk, and it is zero afterwards.

(a) When the women walk with angular velocity **ω
**= -(v/r)** k **= -((1.5 m/s)/(2 m))** k **= -(0.75/s)** k**

her angular momentum is
**L **= I**ω **= (-60 kg
* 4 m^{2}
* 0.75/s)**k **= -(180 kgm^{2}/s)**k**.

The angular momentum of the turntable will be **L **= (180 kgm^{2}/s)**k**

and its angular velocity
**ω **= (180 kgm^{2}/s)**k**/(500 kgm^{2}) = (0.36/s)**k**.

The turntable turns counterclockwise.

(b) The kinetic energy of the women is

½Iω^{2
}= ½ 240 kgm^{2}*(0.75 s)^{2 }= 67.5 J.

The kinetic energy of the turntable is

½Iω^{2
}= ½ 500 kgm^{2}*(0.36 s)^{2 }= 32.4 J.

The work done by the women is W = (67.5 + 32.4)J = 99.9 J.

The total angular momentum about any axis in the universe is conserved. The angular momentum of a single object, however, changes when a net torque acts on the object for a finite time interval. Conversely, if no net torque acts on an object, then its angular momentum is constant.