Work, Energy, and Power in Rotational Motion: How does the shape of the rigid body effects energy of the rotational motion? This article will answer all the questions related to the work, energy, and power for rotating rigid bodies around a fixed axis. “Rotational motion is nothing but the motion of an object around a circular path, in a fixed orbit.” The rotational motion is completely analogous to linear or translational dynamics.

Most of the equations within the mechanics of rotating objects are almost like the equations in linear motion. The rotational kinetic energy of the body doesn’t depend only on its mass; it also depends upon how mass is distributed about the axis of rotation. Some examples of rotational motion about a fixed point in daily life include the rotation of a ceiling fan, the rotation of the minute, and the hour hand in the clock.

Moment of Inertia

The physical quantity that resists the change in its rotational speed is the moment of inertia of a rigid body. It is symbolized as \(I\) and is measured in \({\rm{kg}}\,{{\rm{m}}^2}\)

The moment of inertia is given by:

\(I\, = \,m{r^2}\) where \(m\) is the particle’s mass, and \(r\) is the distance from the axis of rotation.

From the above formula of the moment of inertia, we can say that; the larger the mass, the greater the moment of inertia.

The moment of inertia of a rigid body depends not only on the mass but also on how this mass is distributed about the axis of rotation. Before calculating the moment of inertia of anybody, its distribution must be specified first.

Torque

The rotational equivalent of linear force is known as torque. As linear force is a push or a pull, similarly, the torque is twisting an object about a fixed axis. Mathematically, torque is the product of force and the perpendicular distance from the axis of rotation. It can be written as:

\(\overrightarrow \tau  \, = \,\overrightarrow r \, \times \,\overrightarrow F \)

\( \Rightarrow \overrightarrow \tau  \, = \,\left| {\overrightarrow r } \right|\,\,\left| {\overrightarrow F } \right|\,\sin \,\theta \)

Where \(r\) is the position vector from its axis of rotation, \(F\) is the force, and \(\theta \) is the angle between position vector and force.

From the above expression, we can say that torque will be zero if,

i. Force acting on the body is zero.
ii. The point of application of force lies on the axis of rotation \((\overrightarrow r  = \,0)\)

For a rigid body, torque will also be zero if the force is parallel or intersecting with the axis of rotation.

Work Done in Rotational Motion

The figure below shows a rigid body that has rotated through an equivalent angle \(d\theta \) from  \(A\) to \(B\) under force \(\overrightarrow F \) The external force \(\overrightarrow F \)  is applied to point \(P\) whose position vector is \(\overrightarrow r \) The rigid body is constrained to rotate about a fixed axis perpendicular to the page. It passes through \(O\) Hence, the vector \(\overrightarrow r \)  moves in a circle of fixed radius \(r\) and the vector  \(\overrightarrow {ds} \) will be perpendicular to \(\overrightarrow r \)

Thus,

\(\overrightarrow {ds} \, = \,\overrightarrow {d\theta } \,\, \times \,\overrightarrow r \)

By using the general equation of work done, we obtain:

\(W = \int {\overrightarrow F } .\,\overrightarrow {ds} \)

\( \Rightarrow \int {\overrightarrow F } .(\overrightarrow {d\theta } \, \times \overrightarrow r ) = \int {\overrightarrow {d\theta } \,(\overrightarrow r } \, \times \,\overrightarrow F )\)

where we used the identity \(\overrightarrow a .(\overrightarrow b \,\, \times \,\overrightarrow c ) = \overrightarrow b \,.\,(\overrightarrow c \, \times \,\overrightarrow a )\)

We know that \((\overrightarrow r \,\, \times \,\overrightarrow F )\, = \,\overrightarrow \tau  ,\) by substituting thin in the final expression for the work done on a rigid body:

\(W\, = \,\int {\overrightarrow \tau  \,.\,\overrightarrow {d\theta } } \)

The total work done on the rigid body can be obtained by integrating all the torques over the angle through which the body rotates.

The work done on the rigid body is the dot product of torque and angular displacement. It will be zero if the torque acting on the rigid body is zero or perpendicular to the angular displacement. All particles rotate through the same angle in a rigid body; thus, the work of every external force is equal to the torque times the common incremental angle \(d\theta \). The quantity \(\sum {_i{\tau _i}} \)  is the net torque on the body due to external forces.

Rotational Kinetic Energy

In the figure below, a rigid body rotates in the \(XY\) plane about the \(Z\)- axis, with uniform angular velocity \(\omega \)

Let us consider that the body is made up of a large number of point masses. Let’s assume it consists of masses \({m_1},{m_2},{m_3}……\) at a perpendicular distance \({r_1},\,{r_2},\,{r_3}………\) respectively from the axis of rotation. 

If \({v_1},\,{v_2},\,{v_3}………\) are the linear velocity of the particle then, \({v_1} = {r_1}\omega ,{v_2} = {r_2}\omega ,{v_3} = {r_3}\omega \), and so on.

The kinetic energy of the particle of mass \({m_1}\) is given by,

\(\frac{1}{2}{m_1}{v_1}^2 = \frac{1}{2}{m_1}{r_1}^2\omega \)

Similarly, the kinetic energy of the other particles of the body are \(\frac{1}{2}{m_2}{r_2}^2{\omega ^2},\,\,\frac{1}{2}{m_3}{r_3}^2{\omega ^2}\)

Then the total kinetic energy of the body due to pure rotation of the body will be given by,

\(KE = \frac{1}{2}{m_1}{r_1}^2{\omega ^2} + \,\,\frac{1}{2}{m_2}{r_2}^2{\omega ^2} + \frac{1}{2}{m_3}{r_3}^2{\omega ^2}…….\)

\( \Rightarrow KE = {\left( {\frac{1}{2}{m_1}{r_1}^2 + \frac{1}{2}{m_2}{r_2}^2 + \frac{1}{2}{m_3}{r_3}^2…..} \right)^2}{\omega ^2}\)

\( \Rightarrow KE = \frac{1}{2}\left( {\sum\limits_{i = 1}^n {{m_i}} r_i^2} \right){\omega ^2}\)

We know that \(\left( {\sum\limits_{i = 1}^n {{m_i}} r_i^2} \right) = I\) where \(I\) is the moment of inertia.

\(KE\, = \,\frac{1}{2}I{\omega ^2}\)

Power in Rotational Motion

Derivation of power in rotational motion can be done similarly as it is derived in linear motion. When the force is a constant, then power in the linear motion is given by \(P\, = \,\overrightarrow F \,.\,\overrightarrow v \). In the following discussion, the net torque will always be constant. The instantaneous power or power is defined as the rate at which work is being done,

\(P\, = \,\frac{{dW}}{{dt}}\)

If we have a constant net torque, it becomes \(W = \,\tau \theta ,\) and the power is

\(P\, = \frac{{dW}}{{dt}} = \frac{{d(\tau \theta )}}{{dt}} = \tau \frac{{d\theta }}{{dt}}\)

\( \Rightarrow \,P\, = \,\tau \omega \)

Summary

The dynamics of rotational motion are completely analogs to linear motion. Torque on any point on the rigid body is given by \(\overrightarrow \tau  \, = \,\overrightarrow \tau   \times \overrightarrow F ,\) where \(r\) is the position vector on force from the axis of rotation. The total work done to rotate a rigid body through an angle \(\theta \) about a fixed axis is given by, \(W = \,\int {\overrightarrow \tau  .\overrightarrow {d\theta } } \)

The rotational kinetic energy of the rigid body is given by \(K = \frac{1}{2}I{\omega ^2},\) where \(I\) is the moment of inertia. The power delivered to a system rotating about a fixed axis is the torque times the angular velocity, \(P = \tau \omega \).

Solved Examples On Work, Energy, and Power in Rotational Motion

Q.1. A round grindstone with a moment of inertia \(1600\,{\rm{kg}}\,{{\rm{m}}^2}\) rotates at an angular velocity of \(6\;{\rm{rad}}\,{\rm{ s}}^{ – 1}\). What is grindstone’s rotational kinetic energy?
Ans: Given,
Moment of inertia, \(I\, = \,1600\,{\rm{kg}}\,{{\rm{m}}^2}\)
Angular velocity, \(\omega \, = 6\;{\rm{rad}}\,{\rm{s}}^{ – 1}\)
From the rotational kinetic energy formula,
\({K_R} = \frac{1}{2}I{\omega ^2}\)
\({K_R} = \frac{1}{2} \times 1600 \times {6^2}\)
Hence, \(28800\,{\rm{J}}\) is the rotational kinetic energy of the grindstone.

Q.2. A boat engine operating a \({\rm{9}}{\rm{.0}}\, \times {10^4}\,{\rm{W}}\) is running at \(300\;{\rm{revs}}/{\rm{min}}\). What is the torque on the propeller shaft?
Ans: Given,
Power, \(P = 9.0 \times {10^4}\;{\rm{W}}\)
Angular velocity, \(\omega \,{\rm{ = }}\,{\rm{300}}\;{\rm{rev}}\,{\rm{/}}\,{\rm{min}}\)
Converting angular velocity from \({\rm{rev}}/{\rm{min}}\) to \({\rm{rad}}/{\rm{s}}\)
\(300\;{\rm{ rev mi}}{{\rm{n}}^{ – 1}} = \frac{{300}}{{60}} \times 2 \times 3.14 = 31.4\,{\rm{rad}}\,{\rm{s}^{ – 1}}\)
We know that the power consumed in rotational motion
\(P = \tau \omega \)
\(\tau = \frac{P}{\omega } = \frac{{9.0 \times {{10}^4}\;{\rm{N}}\;{\rm{m}}{{\rm{s}}^{ – 1}}}}{{31.4\,{\rm{rad}}\,{{\rm{s}}^{ – 1}}}} = 2866.2\;{\rm{N}}\,{\rm{m}}\)

FAQs

Q.1. What are the examples of rotational motion about a fixed axis?
Ans: Some of the examples of rotation about a fixed point are,
i. Ceiling fan rotation,
ii. Rotation of the minute and the hour hand in the clock
iii. Opening and closing of doors and windows

Q.2. What is the moment of inertia?
Ans: The moment of inertia measures the object’s resistance to the change in its rotational motion.
For the point mass, the moment of inertia is given by,
\(I = m{r^2}\) where \(m\) is the particle’s mass, and \(r\) is the distance from the axis of rotation.

Q.3. What is torque?
Ans: The twisting effect of the force applied to a rotating object is called torque. It can be obtained by taking the cross product of position vector and force vector.

Q.4. Define tangential acceleration and gives the relation between tangential acceleration and angular acceleration.
Ans: The linear acceleration of a rotating object in the tangential direction of motion is the tangential acceleration. It changes the rotating speed of the object. The relation between tangential and angular acceleration \((\alpha )\) is given by, \({a_1} = r\alpha \) The SI unit of tangential acceleration is \({\rm{m}}/{{\rm{s}}^2}\)

Q.5. What is the formula of rotational kinetic energy?
Ans: The rotational kinetic energy of the rigid body is given by \(K = \,\frac{1}{2}I{\omega ^2}\) Where \(I\) is the moment of inertia and \(\omega \) is the angular velocity.