Work done by Couple, Kinetic Energy of Rotating and Rolling Body and Acceleration of Rolling Body on an Inclined Plane

Work Done by a Couple

Two equal and opposite parallel forces acting on a rigid body at different points constitute a couple. Consider a wheel of radius r rotating about its centre O. Suppose two equal and opposite force F and tangentially at points A and B of a diameter of the wheel as shown in the figure. Let Ï´ be the angle of rotation of the wheel. The couple or torque due to two forces is then constant.

\begin{align*} \text {The work done by each force} = \text {Force} \times \text {Distance} = F \times s = F \times r\theta = F r \theta \\ \text {where} \theta \text { is in radian and distance} = AA’ = BB’ = s= r\theta \\ \therefore \text {Total work done, W } &= Fr\theta + Fr\theta \\ &= 2Fr\theta \dots (i) \\ \text {But torque,} \tau &= F\times 2r = 2Fr \dots (ii) \\ \text {From equations} (ii) \text {and} (iii), \text {we get} \\ \text {Work done by couple, W} = \tau \theta \dots (iii) \end{align*}

Therefore, the work done by a couple is the product of torque and the angle of rotation of rigid body. As the power P is the rate of doing work, then

\begin{align*} p &= \frac {dW}{dt} = \frac {d}{dt} (\tau \theta ) \\ &= \tau \frac {d\theta }{dt} = \tau \omega \\ \therefore \text {Power,} P &= \tau \omega \\ \end{align*}

Comparing with P = F.v, in translation motion, we conclude that torque in rotational motion plays the same role as force does in translational motion.

Kinetic Energy of a Rotating Body

Let us suppose that a body rotating about an axis YY’ under the action of a constant torque(). Suppose the body consists of n particles of masses m₁, m₂, m₃, …., m_n at distance r₁, r₂, r₃, ….. r_n from the axis of rotation respectively. Suppose the body be rotating with uniform angular velocity ω, about their axis. Although each particle within the body has the same angular velocity ω, their linear velocities will be different.

Let v₁, v₂, v₃, …, v_n be the linear velocities of the particles of masses m₁, m₂, m₃, …., m_n respectively. Then

\begin{align*} v_1 = \omega r_1, \: v_2 =\omega r_2, \dots , v_n =\omega r_n \\ \therefore \text {Rotational kinetic energy of the particle of mass} m_1 = \frac 12 m_1v_1^2 \\ &= \frac 12 m_1 (\omega r_1)^2 \\ &= \frac 12 m_1r_1^2\omega ^2 \\ \text {Similarly, the rotational kinetic energy of particles of masses} m_2, m_3, \dots \text {are} \frac 12 m_2r_2^2 \omega ^2 , \frac 12 m_3r_3^2 \omega ^2 \dots \text {respectively} \\ \end{align*}

When the rigid body rotates about the axis, it possesses of rotational kinetic energy. The rotational kinetic K.E. of the rigid body is given by the sum of the kinetic energy of various constituent particles.

\begin{align*} \text {Rotational K.E. of the body, K.E.} &= \frac 12 m_1r_1^2 \omega ^2 + \frac 12 m_2r_2^2 \omega ^2 + \frac 12 m_3r_3^2 \omega ^2 \dots \\ &= \frac 12 (m_1r_1^2 + m_2r_2^2 + m_3r_3^2 + \dots ) \omega ^2 \\ &= \frac 12 (\sum m_ir_i^2) \omega^2 \\ \frac 12 I\omega ^2 \\ \text {where} \: I = \sum m_ir_i^2, \text {moment of inertia of the body about the axis YY’.} \\ \therefore \text {Rotational K.E of the body} &= \frac 12 I\omega ^2 \\ \end{align*}

So, the relational kinetic energy of a body is equal to the half the product of the moment of inertia of the body and the square of the angular velocity of the body about the given axis of rotation.

Kinetic Energy of a rolling Body

Consider a body such as the wheel of mass m and radius R rolling along a straight line on a horizontal plane surface without slipping as shown in the figure. When the body rolls, it rotates about the horizontal axis through the centre of mass and undergoes displacement in a forward direction. So, the body possess both rotational and translational motion.

Let v be the velocity of the centre of mass and time T is the period of rotation of the body. During this time, it describes an angle of 2p radians about the axis. If the body covers a linear distance x in one revolution, the distance covered in one revolution is

\begin{align*} x &= 2\pi R \text {The angular velocity of the body is} \\ \omega &= \frac {2\pi }{T} \\ \text {and the velocity of translation of its centre its centre of mass is } \\ v &= \frac {2\pi R}{T} \\ \text {Comparing these two equations, we get} \\ v &= \omega R \\ \therefore \text {Kinetic energy of rotation, } E_{rot} &= \frac 12 I\omega ^2 \\ \text {and kinetic energy of translation,} E_{trans} &= \frac 12 mv^2 \\ \text {The total K.E. of the rolling body is given by} \\ E &= E_{rot} + E_{trans} \\ &= \frac 12 I\omega ^2 + \frac 12 mv^2 \dots (i) \\ \text {We have,} I = mK^2 , \text {where K is the radius of gyration of the body about the axis through its centre of mass. Then the equation } (i) can be written as \\ E &= \frac 12 mv^2 + \frac 12 m k^2 \omega ^2 \\ &= \frac 12 mv^2 + \frac 12 m K^2 \left ( \frac vR \right ) ^2 \\ \therefore \text {Total K.E., E} &= \frac 12 mv^2 \left ( 1 + \frac {K^2}{R^2} \right ) \\ \end{align*}

Acceleration of a Rolling Body on an Inclined Plane

Consider a body of circular symmetry e.g. a sphere, disc etc. of mass m and radius R, rolling down along a plane inclined to the horizontal at an angle Ï´ as shown in the figure.

If v be the linear velocity acquired by the body on covering a distance s along the plane, it descends through a vertical height h and loses potential energy.

$$\text {Potential energy lost by the body} = mgh$$

This must obviously be equal to the kinetic energy gained by the body.

\begin{align*} \therefore \text {Total K.E. gained by the body} &= \frac 12 mv^2 \left ( 1 + \frac {K^2}{R^2} \right ) \\ \text {As no slipping occurs, mechanical energy is conserved.} \\ \text {So, the loss in potential energy} &= \text {Gain in K.E.} \\ mgh &= \frac 12 mv^2 \left ( 1 + \frac {K^2}{R^2} \right ) \\ \text {or,} \: v^2 &= \frac {2gh}{ \left ( 1 + \frac {K^2}{R^2} \right ) \dots (i) \\ \frac {we have} \\ \sin \theta &= \frac hs \\ \text {or,} \: &= s\sin \theta \\ \text {Substituting the value of h in equation} \: (i) \text {we get} \\ v^2 &= \frac {2gs \sin \theta}{\left ( 1 + \frac {K^2}{R^2} \right )} \dots (ii)\\ \text {When initial velocity,} u=0, \text {for a body starting from rest, the equation of motion for the body will be} \: v^2 = 2as, \text {where a is the linear acceleration of the rolling body. Then, equation} \: (ii) \text {becomes} \\ 2as &= \frac {2gs \sin \theta}{\left ( 1 + \frac {K^2}{R^2} \right )} \\ \therefore a &= \frac {g \sin \theta}{\left ( 1 + \frac {K^2}{R^2} \right)} \\ \text{This is general expression for acceleration of a body rolling down an inclined plane. The above expression can be developed in terms of mass. The total kinetic energy of the rolling object in terms of mass, } \\ K.E. &= \frac 12 mv^2 + \frac 12 I\omega ^2 = \frac 12 v^2 \left ( m + \frac {I^2}{r^2} \right ) \\ \text {As the loss of potential energy} = \text {the gain in kinetic energy,} \\ mgh &= \frac 12 v^2 \left ( m + \frac {I^2}{r^2} \right ) \\ \text {or,} \: mgs \sin \theta &= \frac 12 v^2 \left ( m + \frac {I^2}{r^2} \right ) \\ \text {or,} \: v^2 &= \frac {2mg s\sin \theta }{M + (I/r^2) } \\ \text {But} \: v^2 = 2 as, \text {where a is the linear acceleration down the plane, then} \\ 2as &= \frac {2mg s\sin \theta }{M + (I/r^2) } \\ \text {or,} \: a &= \frac {mg \sin \theta }{M + (I/r^2) } \end{align*}

Analogue between Translational and Rotational Motions