The expressions of work in pure translation in x- direction (one dimensional linear motion) is given by :

đ W = F x đ x ⇒ W = ∫ F x đ x đ W = F x đ x ⇒ W = ∫ F x đ x

The corresponding expressions of work in pure rotation (one dimensional rotational motion) is given by :

Obviously, torque replaces component of force in the direction of displacement (Fx) and angular displacement (θ) replaces linear displacement (x). Similarly, the expression of kinetic energy as derived earlier in the course is :

Here, moment of inertia (I) replaces linear inertia (m) and angular speed (ω) replaces linear speed (v). We, however, do not generally define gravitational potential energy for rotation as there is no overall change in the position (elevation) of the COM of the body in pure rotation. This is particularly the case when axis of rotation passes through COM.

This understanding of correspondence of expressions helps us to remember expressions of rotational motion, but it does not provide the insight into the angular quantities used to describe pure rotation. For example, we can not understand how torque accomplishes work on a rigid body, which does not have any linear displacement ! For this reason, we shall develop expressions of work by considering rotation of a particle or particle like object in the next section.

We need to emphasize here another important underlying concept. The description of motion in pure translation for a single particle and a rigid body differ in one very important aspect. Recall that the motion of a rigid body, in translation, is described by assigning motional quantities to the "center of mass(COM)". If we look at the expression of center of mass in x-direction, then we realize that the concept of COM is actually designed to incorporate distribution of mass in the body.

We must note here that such distinction arising due to distribution of mass between a particle and rigid body does not exist for pure rotation. It is so because the effect of the distribution of mass is incorporated in the definition of moment of inertia itself :

For this reason, there is no corresponding "center of rotational inertia" or "center of moment of inertia" for rigid body in rotation. This is one aspect in which there is no correspondence between two motion types. It follows, then, that the expressions of various quantities of a particle or a rigid body in rotation about a fixed axis should be same. Same expressions would, therefore, determine work and kinetic energy of a particle or a rigid body. The difference in two cases will solely arise from the differences in the values of moments of inertia.

We consider a particle like object, attached to a "mass-less" rod, to simulate rotation of a particle. The particle like object rotates in a plane perpendicular to the plane of screen or paper as shown in the figure. Here, we shall evaluate work by a force, "F", in xy-plane, which is perpendicular to the axis of rotation.

We know that it is only the tangential component of force that accelerates the particle and does the work. The component of force perpendicular to displacement does not perform work. For small linear displacement "ds", we have :

Figure 3: A particle like object in rotation.

Rotation

đ W = F x đ x = F t đ s đ W = F x đ x = F t đ s

By geometry,

= r đ θ = r đ θ

Combining two equations, we have :

đ W = F t r đ θ = τ đ θ đ W = F t r đ θ = τ đ θ

Work done by a torque in rotating the particle like body by an angular displacement, we have :

⇒ W = ∫ τ đ θ ⇒ W = ∫ τ đ θ

If work is done by a constant torque, then we can take torque out of the integral sign and :

⇒ W = τ ∫ đ θ = τ Δ θ ⇒ W = τ ∫ đ θ = τ Δ θ

As discussed earlier, this same expressions is valid for work done on a rigid body in pure rotation.

Power in rotation

Power measures the rate at which external torque does work on a particle like object or a rigid body. It is equal to the first time differential of work done,

P = đ W đ t = τ đ θ đ t = τ ω P = đ W đ t = τ đ θ đ t = τ ω

(5)

We must note the correspondence of this expression with its linear counterpart, where :

P = F v P = F v

Clearly, torque (τ) replaces force (F) and angular speed (ω) replaces linear speed (v) in the expression of power.

Example 1

Problem : A uniform disk of mass 5 kg and radius 0.2 m is mounted on a horizontal axle to rotate freely. One end of a rope, attached to the disk, is wrapped around the rim of the disk. If a force of 10 N is applied vertically as shown, then find the work done by the force in rotating the disk for 2 s, starting from rest at t = 0 s.

Figure 4: A uniform disk is mounted on a horizontal axle.

Rotation

Solution : Here, torque due to vertical force is constant. Thus, in order to determine the work on the disk in rotation, we need to use the expression of work done by a constant torque as :

W = τ Δ θ = τ ( θ 2 - θ 1 ) W = τ Δ θ = τ ( θ 2 - θ 1 )

It means that we need to calculate torque and angular displacement. Here, torque (positive as rotation is anti-clockwise) is given by :

τ = F t r = 10 x 0.2 = 2 N-m τ = F t r = 10 x 0.2 = 2 N-m

This is a rotational motion with constant acceleration (due to constant torque). Thus, we can use the equation of motion for constant angular acceleration to find the displacement :

θ f = θ i t + 1 2 α t 2 θ f = θ i t + 1 2 α t 2

Here, disk is at rest in the beginning, θi = 0. Hence,

⇒ θ f = 1 2 α t 2 ⇒ θ f = 1 2 α t 2

But, we do not know the angular acceleration. We can, however, use corresponding Newton's second law for rotation. Since, disk rotates freely, we can assume that there is no other torque involved :

⇒ α = τ I ⇒ α = τ I

On the other hand, the moment of inertia for the disk about an axis passing through the center of mass and perpendicular to its surface is given by :

I = M R 2 2 = 5 x 0.2 2 2 = 0.1 kg - m 2 I = M R 2 2 = 5 x 0.2 2 2 = 0.1 kg - m 2

Putting this value of moment of inertia, we get the value of angular acceleration :

⇒ α = τ I = 2 0.1 = 20 rad s 2 ⇒ α = τ I = 2 0.1 = 20 rad s 2

Now final angular position is :

θ f = 1 2 x 20 x 2 2 = 40 rad θ f = 1 2 x 20 x 2 2 = 40 rad

Hence, work done is :

W = τ ( θ 2 - θ 1 ) = τ θ 2 = 2 x 40 = 80 J W = τ ( θ 2 - θ 1 ) = τ θ 2 = 2 x 40 = 80 J

We have seen earlier that net work on a body is equal to change in its kinetic energy.

In translation, we have :

W = Δ K = 1 2 m v f 2 - 1 2 m v i 2 W = Δ K = 1 2 m v f 2 - 1 2 m v i 2

A corresponding work - kinetic energy theorem for pure rotation, therefore, is :

where "W" represents the net work on the rigid body in rotation.

Example 2

Problem : A uniform disk of mass 5 kg and radius 0.2 m is mounted on a horizontal axle to rotate freely. One end of a rope, attached to the disk, is wrapped around the rim of the disk. If a force of 10 N is applied vertically as shown, then find the work done by the force in rotating the disk for 2 s, starting from rest at t = 0 s. Use work - kinetic energy theorem to find the result.

Figure 5: A uniform disk is mounted on a horizontal axle.

Rotation

Solution : This is the same question as the previous one except that we are required to solve the question, using work-kinetic energy theorem :

W = Δ K = 1 2 I ω f 2 - 1 2 I ω i 2 W = Δ K = 1 2 I ω f 2 - 1 2 I ω i 2

From the previous example, we have :

I = M R 2 2 = 5 x 0.2 2 2 = 0.1 kg - m 2 I = M R 2 2 = 5 x 0.2 2 2 = 0.1 kg - m 2

and

α = τ I = 20 rad s 2 α = τ I = 20 rad s 2

Now, as the disk starts from rest, ω i = 0 ω i = 0 . In order to find the final angular speed, we use equation of motion for constant angular acceleration :

ω f = ω i + α t = 0 = 2 x 20 = 40 rad / s ω f = ω i + α t = 0 = 2 x 20 = 40 rad / s

Thus,

W = Δ K = 1 2 I ω f 2 - 0 = 1 2 I ω f 2 ⇒ W = 1 2 x 0.1 x 40 2 = 80 J W = Δ K = 1 2 I ω f 2 - 0 = 1 2 I ω f 2 ⇒ W = 1 2 x 0.1 x 40 2 = 80 J

As expected, the result is same as calculated in earlier example.

1: Rotational quantities describing the motion are linear (one-dimensional) vectors.

2: Work done by a torque is given by :

W = ∫ τ đ θ W = ∫ τ đ θ

3: Work done by a constant torque is given by :

W = τ Δ θ W = τ Δ θ

4: We generally do not consider change in gravitational potential energy in the case of rotation about an axis that passes through its center of mass. Since there is no change in the position of center of mass, there is no corresponding change in its potential energy.