## Rotational dynamics

We have defined the angular displacement, angular speed and angular velocity, angular acceleration, and kinetic energy of an object rotating about an axis.  These definitions apply to objects spinning about an internal axis, such as a wheel spinning on its axle, or to objects revolving around a point external to the objects, such as the earth revolving around the sun.
A spinning or revolving object has angular velocity ω.  Whenever the magnitude or direction of this angular velocity changes, the object has angular acceleration α.

### What causes angular acceleration?

Assume you want to change a rotating wheel's angular speed.  To increase the angular speed you probably will apply a force to the rim, tangential to the rim, and in the direction of the instantaneous velocity of the section of the rim to which you apply the force.

If you want to decrease the angular speed, you will reverse the direction of the force.

Assume you want to enter a building with a rotating door.  The door has four panels, and you push on one of them, perpendicular to the surface of the panel.
The rate, at which the angular velocity of the door changes, i.e. the angular acceleration α, is greater the farther away from the axis of rotation you apply the force.

Angular acceleration about a point is the result of a torque about this point.  A torque is the product of a lever arm and a force that is applied perpendicular to the lever arm.  The lever arm or moment arm is the perpendicular distance from the center of rotation, i.e. from the pivot point, to the point where the force is applied.

A torque is always defined with respect to a pivot point.

A larger torque produces a larger angular acceleration.  You can get a larger torque by applying a larger force, or by using a longer lever arm.  We write

torque = lever arm × force,

Torque is a vector.  It is the vector product or cross product of r and F.
If a force F acts on an object, then the torque produced by this force about a pivot point is τ = r × F, where r is a displacement vector pointing from the axis of rotation to the point where the force is applied.
The SI units of torque are Nm.
Torque has magnitude and direction.
Its direction is given by the right-hand rule.

Let the fingers of your right hand point from the axis of rotation to the point where the force is applied.
Curl them into the direction of F
Your thumb points in the direction of the torque vector.

The magnitude of the torque τ is  τ = rFsinθ, where θ is the smallest angle between the directions of the vectors r and F
We can also write τ = rperpF = rFperp, where  rperp is the component of the lever arm perpendicular to F, or where Fperp is the component of F perpendicular to the lever arm.

If two or more forces act on an object, then the net torque is the vector sum of the torques produced by the separate forces.  (For rotations about a single axis, two torques can point in the same direction or in opposite directions and therefore can add or subtract.)

#### Problem:

In the figure on the right, find the net torque on the wheel about the axle through the center if a = 10 cm and b = 25 cm.

Solution:
Let the z-axis come out of the page.
The torque produced by the 10 N force then is τ = -(10 N  * 0.25 m) k = -(2.5 Nm) k.
The torque produced by the 9 N force is τ = -(9 N * 0.25 m) k = -(2.25 Nm) k.
The torque produced by the 12 N force is τ = (12 N * 0.1 m) k = (1.2 Nm) k.
(Note that this force is also applied perpendicular to the lever arm.)
The total torque is τ = -(3.55 Nm) k.

#### Problem:

Given M = 6i + 2j - k and N = 2i - j - 3k calculate the vector product M × N.

Solution:
× N = (6i + 2j - k) × (2i - j -3k)
= 12i × i - 6i × j - 18i × k + 4j × i - 2j × j - 6j × k - 2k × i + k × j + 3k × k
= 0 - 6k +18j - 4k - 0 - 6i - 2j - i + 0
= -7i + 16j - 10k.

Newton's second law, when applied to rotational motion states that the torque equals the product of the rotational mass or moment of inertia I and the angular acceleration α.

torque = moment of inertia × angular acceleration
τ = Iα

The moment of inertia of an object is a measure of its resistance to angular acceleration.  Because of its rotational inertia you need a torque to change the angular velocity of an object.  If there is no net torque acting on an object, its angular velocity will not change.  If it is initially not spinning, it will not start spinning.  If it is spinning with a given angular velocity, this angular velocity will not change.  Both, its angular speed and the orientation of its axis of rotation, will stay the same.

constant angular velocity <--> no angular acceleration <--> no net torque

When two objects are acted on by the same torque, the object with the larger moment of inertia has the smaller angular acceleration.

### Work and energy in rotational motion

Assume a particle is displaced a distance rdθ along a circular path.  If a tangential force acts on the particle through the displacement, then the work done by this force is W = Frdθ = rFdθ = τdθ.

W = τdθ

The rate at which work is done (i.e. the power) is P = dW/dt = τdθ/dt = τω.

P = τω.

The net work done on a rotating object is equal to the change in the rotational kinetic energy of the object.

#### Example:

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 = ½Iω2.  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.)

#### Problem:

A model airplane the mass of which is 0.75 kg is tethered by a wire so that it flies in a circle 30 m in radius.  The airplane engine provides a net thrust of 0.8 N perpendicular to the tethering wire.
(a)  Find the torque that the net thrust produces about the center of the circle.
(b)  Find the angular acceleration of the airplane when it is in level flight.
(c)  Find the linear acceleration of the airplane tangent to its flight path.

Solution:
(a)  Let the circle lie in the x-y plane and the z-axis point out of the page.
Then τ = r × F = (0.8 N * 30 m)k = 24 Nmk.
(b)  τ = Iα.  I = mr2 = (0.75 kg)(30 m)2 = 675 kgm2.  α = τ/I = 0.036/s2k.
(c)  a = αr = (0.036/s2)(30 m) = 1.07 m/s2.

### The seesaw

Two children are playing on a seesaw, rocking back and forth.  The center of the seesaw is fixed.  There is no translational motion.  We observe rotational motion about the center.  While the seesaw is moving there is practically no torque on the seesaw, and it rotates with uniform angular velocity.  The weight of each child times the lever arm from the center to where the child is seated produces a torque, but the two torques have the same magnitude and point in opposite directions and therefore cancel.  If one child is heavier, it sits closer to the center than the lighter child.  This reduces the lever arm.  In this way, different weights can produce torques of the same magnitude.

When one child's feet hit the floor, the floor pushes back and produces a torque pointing opposite to the angular velocity.  This torque will reduce the angular velocity to zero in a short time interval.  The seesaw is now stopped.  The child then pushes off and the ground pushes back.  The direction of the torque stays the same.  The torque produces an angular acceleration which results in an angular velocity pointing opposite to the original direction.  The seesaw reaches its final angular velocity when the child stops pushing.  It now keeps on rotating with constant angular velocity, until the other child's feet hit the floor.