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P = F v

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

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.

Rotation

A uniform disk is mounted on a horizontal axle.

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 )

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

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

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

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

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

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

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

Hence, work done is :

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

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Work - kinetic energy theorem for rotational motion

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

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

W = Δ K = 1 2 I ω f 2 - 1 2 I ω i 2

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

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.

Rotation

A uniform disk is mounted on a horizontal axle.

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

From the previous example, we have :

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

and

α = τ I = 20 rad s 2

Now, as the disk starts from rest, 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

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

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

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Summary

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

2: Work done by a torque is given by :

W = τ đ θ

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

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.

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Source:  OpenStax, Physics for k-12. OpenStax CNX. Sep 07, 2009 Download for free at http://cnx.org/content/col10322/1.175
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