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 w.  Whenever the magnitude or direction of this angular velocity changes, the object has angular acceleration a.

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,

or

t = r ´ F.

Torque is a vector.  It is the vector product or cross product of r and F.  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 vector product of two vectors A and B is defined as the vector C=A´B.

The magnitude of C is C = ABsinq, where q is the smallest angle between the directions of the vectors A and B.  C is perpendicular to both A and B, i.e. it is perpendicular to the plane that contains both A and B.  The direction of C can be found by using the right-hand rule.
Let the fingers of your right hand point in the direction of A.  Orient the palm of your hand so that, as you curl your fingers, you can sweep them over to point in the direction of B.  Your thumb points in the direction of C = A´B.
If A and B are parallel or anti-parallel to each other, then C = A´B = 0, since sinq = 0.  If A and B are perpendicular to each other, then sinq = 1 and C has its maximum possible magnitude.

We can find the Cartesian components of C = A´B in terms of the components of A and B.

C_x = A_yB_z - A_zB_y
C_y = A_zB_x - A_xB_z
C_z = A_xB_y - A_yB_x 

If a force F acts on an object, then the torque produced by this force about a pivot point is t = r´F, where r is the displacement vector from the pivot point to the point where the force is applied.  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.

Problems:

The angular acceleration of an object is proportional to the torque acting on the object.  If a torque causes angular acceleration about one of the symmetry axes of an object, then the proportional constant is the moment of inertia of the object about the axes.  We write

torque = moment of inertia ´ angular acceleration

or

t = Ia .

The moment of inertia is a measure of an object's rotational inertia.  It depends on the mass of the object, and on how this mass is distributed with respect to the axis of rotation.  The farther the bulk of the mass is from the axis of rotation, the greater is the rotational inertia (moment of inertia) of the object.  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.

Problem:

Equilibrium of a rigid object

An object is said to be in static equilibrium if its linear acceleration as well as its angular acceleration are zero.

Assume two or more forces act on a rigid object.  The total force on the object is the sum of all the external forces.  According to Newton's third law, the sum of all the internal forces is zero.  We have

F_tot = Ma_CM.

If the total force acting on the object is zero, the center of mass of the object will not accelerate.  The object is in translational equilibrium.  If the CM of the object is at rest, it will stay at rest, if it is moving with constant velocity, it will keep on moving with constant velocity.

For the object to be in rotational equilibrium, the total torque on the object must be zero.  The total torque on the object is the vector sum of all the external torques.  According to Newton's third law, the sum of all the internal torques is zero.

The total torque can be nonzero, even though F_tot = 0.

Consider the situation shown in the figure above.  The square is initially at rest in the x-y plane.  The z-axis points out of the page.  The total force on the square is zero.  The CM of the square will stay at rest.  The total torque about the CM is .  The square will have angular acceleration pointing into the page.

Two forces, equal in magnitude and opposite in direction, whose lines of action do not pass through the same center, are called a couple.  A couple produces a torque about the center.

For an object to be in rotational equilibrium and to not have angular acceleration, we need the total torque acting on the object to be zero.  The conditions for static equilibrium therefore are

F_tot = 0, t_tot = 0.

Problem:

When an object is placed in a uniform gravitational field, then the force of gravity produces no torque about the center of mass of the object.  We can always represent the gravitational force as an arrow pointing straight towards or away from the center of mass of the object.  For this reason, the center of mass is also called the center of gravity of the object.

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.

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