Rolling Motion: Elaboration, Formula, and Derivation

Rolling Motion: In our day-to-day life, we observe the various rolling objects such as wheels of a car moving or wheels attached to your bicycle. Do you know the kind of motions that an object and its particles undergo while in rolling motion? The rolling motion is the combination of rotational and translational motion. The velocity at any point on the rigid rolling body depends on the radial distance of a point from the rotating axis. In this topic, we will discuss rolling motion, rolling motion without slipping, rolling motion with slipping, and also kinetic energy of rolling motion.

Rolling Motion

The rolling motion is the most common motion observed in daily life. For example, we shall begin with the case of a disc, but the result will apply to any rolling body that rolls on a level surface. Let us assume that the disc rolls without slipping. To understand, let us suppose a disc that has a translational velocity \(\left( {{v_{{\rm{cm}}}}} \right)\) parallel to the level surface. Since the centre of mass of the rolling disc is at its geometric centre \(C\) as shown in the figure below.

The disc rotates about its symmetrical axis, which passes through \(C.\) Thus, we can write the velocity at any point of the disc, like \({P_0},\;{P_1},{P_2}\) consists of two parts; one is the translational velocity \(\left( {{v_{{\rm{cm}}}}} \right),\) and the other is the linear velocity \(\left( {{v_{\rm{r}}}} \right)\) on account of rotation. Where,  \({v_r} = r\omega ,\)where \(\omega \) is the angular velocity of the rotation of the disc about the axis and \(r\) is the distance of the point from the axis (i.e. from \(C\)). The velocity \({v_r}\) is directed perpendicular to the radius vector of the given point to \(C,\) as shown in the figure. The velocity of the point \({P_2}\left( {{v_2}} \right)\) and its components \({v_r}\) and \(\;{v_{{\rm{cm}}}}\) are shown; \(\;{v_r}\) here is perpendicular to \(C{P_2}.\) It is easy to show that \({v_Z}\) is perpendicular to the line \({P_0}{P_2}.\) Therefore the line passing through \({P_0}\) and parallel to \(\;\omega\)is called the instantaneous axis of rotation. At \({P_0}\) the linear velocity \(\;{v_r}\) due to rotation is directed exactly opposite to the translational velocity \(\left( {{v_{{\rm{cm}}}}} \right).\) Similarly, the magnitude of \({v_r}\) will be \(R\;\omega ,\) where the radius of the disc is \(R.\) The condition that \({P_0}\) is instantaneously at rest requires \({v_{{\rm{cm}}}} = R\omega .\)

Moment of Inertia

For a rigid body, the moment of inertia is defined a the resistance to the change in its rotation. The moment inertia is symbolized as \(I\) and is

in \({\rm{kg}}\,{\rm{m}}^{2}.\)
The moment of inertia is given by:
\(I = m{r^2}\) where the mass of the body \( = m,\) and the distance from the axis of rotation \(= r.\)

Rolling Motion without Slipping

This situation can be observed in the interaction of a car’s tires and the road surface. When we suddenly apply pressure on the accelerator pedal of the car, then the tires spin without the car moving forward (slipping). But after some time, the car moves forward and the tires roll without slipping.

We can analyze rolling without slipping by deriving the relationship between the linear variables of velocity and acceleration of the centre of mass of the wheel in terms of the angular variables that describe the wheel’s motion. The situation is shown below.

Figure: (a) The force of static friction \(\overrightarrow {{f_s}} ,\left| {\overrightarrow {{f_s}} } \right| \le {\mu _s}N\) is large enough to avoid slipping. (b) The linear velocity and acceleration vectors of the centre of mass and the relevant expressions for \(\omega \) and \(\alpha .\) Point \({{P}}\) is at rest relative to the surface.

In rolling without slipping, point \({{P}}\) that is in contact with the surface is at rest relative to the surface. Point \({{P}}\) has velocity \( – R\omega \hat i\)  with respect to the centre of the wheel.
We know that in rolling without slipping, the velocity of point \({{P}}\) relative to the surface is zero, \(\;{v_p} = 0,\) therefore,
\({v_{{\rm{cm}}}} = R\omega \)
After differentiating the above equation, we obtain an expression for the linear acceleration of the centre of mass, and it is given by,
\({a_{{\rm{cm}}}} = R\alpha\)
Where \(R\) is a constant, \(\alpha  = \frac{{d\omega }}{{dt}},\) and \(a = \frac{{dv}}{{dt}}.\)

When the wheel rolls from point \(A\) to point \(B,\) then the distance travelled is \({d_{{\rm{cm}}}}\).
We can observe from the above figure that the distance travelled is equal to the arc length \(R\theta.\) Then we can write as, \({d_{{\rm{cm}}}} = R\theta\)

Rolling Motion with Slipping

In this case, Point \(P\) is not at rest to the ground and have non zero relative motion with respect to ground. This happens when the friction is not large enough to avoid slipping. The situation is shown below figure. In  the case of slipping, \({v_{{\rm{cm}}}} – R\omega \ne 0,\) and \({v_P} \ne 0.\) Thus, \(\omega  \ne \frac{{{v_{{\rm{cm}}}}}}{R}\) and \(\alpha  \ne \frac{{{a_{{\rm{cm}}}}}}{R}.\)

Kinetic Energy of Rolling Motion

In Rolling motion, the kinetic energy of a system of particles \(\left( K \right)\) can be divided into the transitional kinetic energy of the centre of mass \(\left( {\frac{{m{v^2}}}{2}} \right)\) and rotational kinetic energy about the centre of mass of the system of particles \(\left( {K’} \right).\)
Thus the total kinetic energy, \(K = K’ + \frac{{m{v^2}}}{2}.\)

We know that the transitional kinetic energy of the centre of mass of the rolling body is \(\;\frac{{mv_{{\rm{cm}}}^2}}{2},\) where \(m\) is the mass of the body and \({{v_{{\rm{cm}}}}}\) is the centre of the mass velocity. \(\;K’\) represents the kinetic energy of rotation of the body; \(K’ = \frac{1}{2}I{\omega ^2},\) where \(I\) is the moment of inertia about the rotating axis. The kinetic energy of a rolling body is given by,
\(K = \frac{1}{2}I{\omega ^2} + \frac{1}{2}mv_{{\rm{cm}}}^2\)

We are putting, \(I = m{k^2},\) where \(k\) is the radius of gyration of the body and \({v_{{\rm{cm}}}} = R\omega.\)
We get,
\(\;K = \frac{1}{2}m{k^2}{\left( {\frac{{{v_{{\rm{cm}}}}}}{R}} \right)^2} + \frac{1}{2}mv_{{\rm{cm}}}^2\)
\(K = \frac{1}{2}mv_{{\rm{cm}}}^2\left( {1 + \frac{{{k^2}}}{{{R^2}}}} \right)\)
This equation applies to any rolling body such as a disc, cylinder, ring or sphere.

Solved Examples on Rolling Motion

Q.1. Three bodies, such as a ring, a solid cylinder, and a solid sphere with an identical radius, started from rest and rolled down from the same inclined plane without slipping. Then,  which of the bodies reaches the ground with maximum velocity?
Solution: There is no loss of energy due to friction etc. Hence, using energy conservation, the body loses the potential energy in rolling down the inclined plane. \(\left( { = mgh} \right)\) be equal to kinetic energy gained. As the bodies start from rest, the kinetic energy gained is equal to the final kinetic energy of the bodies.
The total kinetic energy in rolling is given by,
\(K = \frac{1}{2}mv_{{\rm{cm}}}^2\left( {1 + \frac{{{k^2}}}{{{R^2}}}} \right),\)

Here, \(v\) is the final velocity of (the centre of mass of) the body.
Equating \(K\) and \(mgh,\) we have
\(mgh = \frac{1}{2}mv_{{\rm{cm}}}^2\left( {1 + \frac{{{k^2}}}{{{R^2}}}} \right)\)
\(\Rightarrow v_{{\rm{cm}}}^2 = \frac{{2gh}}{{\left( {1 + \frac{{{k^2}}}{{{R^2}}}} \right)}}\)
Note \({v_{{\rm{cm}}}}\) is independent of the mass of the rolling body.
For a ring, \({k^2} = {R^2}\)
\({v_{{\rm{ring}}}} = \sqrt {\frac{{2gh}}{{\left( {1 + 1} \right)}}} = \sqrt {gh}\)
For a solid cylinder, \({k^2} = \frac{{{R^2}}}{2}\)
\({v_{{\rm{cylinder}}}} = \sqrt {\frac{{2gh}}{{\left( {1 + \frac{1}{2}} \right)}}} = \sqrt {\frac{{4gh}}{3}}\)
For a solid sphere, \(\;{k^2} = \frac{{2{R^2}}}{5}\)
\({v_{{\rm{sphere}}}} = \sqrt {\frac{{2gh}}{{\left( {1 + \frac{2}{5}} \right)}}} = \sqrt {\frac{{10gh}}{7}}\)
From the results, we obtained that \(\omega \) the sphere has the greatest among the three bodies, and the ring has the least velocity of the centre of mass at the bottom of the inclined plane.

Q.2.  A disc is standing on a flat rough surface. Then, its centre is suddenly given a velocity of  \(5\,{\rm{m}}\,{{\rm{s}}^{{\rm{ – 1}}}}\) in the forward direction. Find the time in which pure rolling will start? Mass \( = {{1}}\,{\rm{kg}},\) Radius \(= 10\,{\rm{cm}}\). Coefficient of friction \( = {\rm{ }}{\bf{0}}.{\bf{4}}.\)
Solution: Friction force will decrease its linear velocity & increase its angular velocity.
\(f = ma\) Friction’ \(f = \mu N = 0.4 \times 1 \times 10\; = 4\;{\rm{N}}\)
Thus, \(v = u–at\) & \(\;\omega = {\omega _0} + \alpha t = {\omega _0} + \frac{a}{R}t\)
For pure rolling: \(v = R\omega \)
Thus, \(u–at = {\omega _0} + \frac{a}{R}t\)
\(\Rightarrow \,\,\,5–4t = 0 + \left( {\frac{4}{{0.1}}} \right)t\)
\(\Rightarrow 5 = 44t\)
\(\Rightarrow \frac{5}{{44}} = t = 0.113\;{\rm{s}}\)

Summary

The rolling motion is a common combination of rotational and translational motion. In rolling without slipping, static friction \(\overrightarrow {{f_s}}\) \(\left| {\overrightarrow {{f_s}} } \right| \le {\mu _s}N\;\) acts on the body to avoid slipping. In rolling without slipping, the point of contact will be at rest to the surface. So, \(\;{v_{{\rm{cm}}}} = R\omega ,\,\,{a_{{\rm{cm}}}} = R\alpha \) and \({d_{{\rm{cm}}}} = R\theta. \) In rolling with slipping, kinetic friction force acts between surfaces, and the point of contact will not be at rest to the surface. So, \({v_{{\rm{cm}}}} \ne R\omega ,\,\,{a_{{\rm{cm}}}} \ne R\alpha \) and \({d_{{\rm{cm}}}} \ne R\theta .\) The total kinetic energy in rolling motion is given by \(\;K = \frac{1}{2}mv_{{\rm{cm}}}^2\left( {1 + \frac{{{k^2}}}{{{R^2}}}} \right),\) where \(k = \) the corresponding radius of gyration of the body.

FAQs on the Rolling Motion

Q.1. What is the relation between angular and linear velocity in pure rolling or rolling without slipping?
Ans: For pure rolling \(v = R\omega,\) at every instant of time, where v is the speed of the centre of the sphere & \(\omega\) is its angular speed.

Q.2. Can a sphere roll purely on a smooth inclined surface?
Ans: For pure rolling \(v = R\omega,\) at every instant of time, (where \(v\) is the speed of the centre of the sphere & \(\omega\) is its angular speed). On an inclined surface, a component of gravity is converting its potential energy to kinetic energy & thus ‘\(v\)’ is increasing, so ‘\(\omega\)’ should also increase. This requires the presence of torque, which in this case, is to be provided by friction. Hence, pure rolling on an incline requires the surface to be rough.

Q.3. What will be kinetic energy in pure rolling or rolling without slipping?
Ans: The total kinetic energy in rolling motion is given by \(\;K = \frac{1}{2}mv_{{\rm{cm}}}^2\left( {1 + \frac{{{k^2}}}{{{R^2}}}} \right),\) where \(k = \) the corresponding radius of gyration of the body.

Q.4. Which type of friction force acts while pure rolling?
Ans: There must be static friction force between both surfaces in pure rolling or rolling without slipping.

Q.5. What will be the kinetic energy in pure rotational motion?
Ans: The kinetic energy in the pure rotational motion is given by \(K = \frac{1}{2}I{\omega ^2},\) where \(\;I\) is the moment of inertia and \(\omega\) is the angular velocity.

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