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**.

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
= r_{perp}F =
rF_{perp}, where r_{perp} is the component of
the lever arm perpendicular to F, or where F_{perp} 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.)

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**.

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

Solution:**M **×** N** = (6**i
**+ 2**j **-** k**) × (2**i **-** j **-3**k**)

= 12**i **×** i
**- 6**i **×** j **- 18**i **×** k **+ 4**j **×** i **- 2**j
**×** j **- 6**j **×** k **- 2**k **×** i **+** k **×** j
**+ 3**k **×** k**

= 0 - 6**k **+18**j
**- 4**k **- 0 - 6**i **- 2**j **-** i **+ 0

= -7**i **+ 16**j
**- 10**k.**

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** = I**α**

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.

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.

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.)

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 Nm**k**.

(b)
**τ** = I**α**. I = mr^{2
}= (0.75 kg)(30 m)^{2 }= 675 kgm^{2}.
** α **=** τ**/I = 0.036/s^{2}k.

(c) a = αr = (0.036/s^{2})(30 m)
= 1.07 m/s^{2}.

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.

Link: Seesaw