Rolling Motion

What Rolling Motion Is

Objects such as a cylinder, a sphere or a wheel are quite often seen to perform rolling motion. In pure rolling, two motions are undertaken simultaneously — circular motion and linear motion. The individual motion of the particles (except the one at the centre of mass) is too difficult to describe, so for theory we consider the actual motion to be the result of

• the rotational motion of the body as a whole about its own symmetric axis, and
• the linear motion of the body assuming its whole mass to be concentrated at its centre of mass.

In other words, the centre of mass performs purely translational motion. Accordingly the object possesses two kinds of kinetic energy, rotational and translational, and the sum of the two is its total kinetic energy.

Static friction is essential for a purely rolling motion — it is what prevents sliding. This is why an object rolling down may later start sliding, and why a sphere pushed along a horizontal surface first slips for some distance and only then starts rolling.

For a body of radius $R$ rolling without slipping, the speed of the centre of mass and the angular speed are related by

$$v = R\omega$$

Derivation: Total Kinetic Energy of a Rolling Body

Let $M$ = mass of the body, $v$ = linear speed of its centre of mass, $R$ = radius, $\omega$ = angular speed of rotation, $I$ = moment of inertia about its own axis and $K$ = radius of gyration, so that $I = MK^2$ and $\omega = \dfrac{v}{R}$.

Total kinetic energy of rolling = translational K.E. + rotational K.E.

$$E = \dfrac{1}{2}Mv^2 + \dfrac{1}{2}I\omega^2 = \dfrac{1}{2}Mv^2 + \dfrac{1}{2}MK^2\left(\dfrac{v^2}{R^2}\right)$$

$$\therefore\ E = \dfrac{1}{2}Mv^2\left[1+\dfrac{K^2}{R^2}\right]$$

Since the value of $K$ differs from body to body, the total kinetic energy of rolling also varies from body to body for the same mass and speed.

For a solid sphere, $I = \dfrac{2}{5}MR^2 = MK^2$, so $\dfrac{K^2}{R^2} = \dfrac{2}{5}$ and

$$E = \dfrac{1}{2}Mv^2\left(1+\dfrac{2}{5}\right) = \dfrac{7}{10}Mv^2$$

Linear Acceleration and Speed While Rolling Down an Inclined Plane

Diagram

Draw an inclined plane making an angle $\theta$ with the horizontal. Show a circular body of mass $M$ and radius $R$ on the incline, rolling down through a vertical height $h$ while travelling a distance $s$ along the plane, so that $h = s\sin\theta$. Show $Mg$ vertically downwards, $N$ perpendicular to the incline, and the static friction $f_s$ up the incline at the point of contact.

Derivation

As the object starts rolling down, its gravitational potential energy is converted into the kinetic energy of rolling. Starting from rest, let $v$ be the speed of the centre of mass after the object has come down through a vertical distance $h$.

$$Mgh = \dfrac{1}{2}Mv^2\left[1+\dfrac{K^2}{R^2}\right]$$

$$\therefore\ v = \sqrt{\dfrac{2gh}{1+\dfrac{K^2}{R^2}}}$$

The linear distance travelled along the plane is $s = \dfrac{h}{\sin\theta}$. During this distance the linear velocity has increased from zero to $v$. If $a$ is the linear acceleration along the plane, then from $v^2 = u^2 + 2as$ with $u = 0$,

$$\dfrac{2gh}{1+\dfrac{K^2}{R^2}} = 2a\left(\dfrac{h}{\sin\theta}\right)$$

$$\therefore\ a = \dfrac{g\sin\theta}{1+\dfrac{K^2}{R^2}}$$

For pure sliding without friction the acceleration is $g\sin\theta$ and the final speed is $\sqrt{2gh}$. Thus during pure rolling the factor $\left(1+\dfrac{K^2}{R^2}\right)$ is effective in both expressions. Since this factor contains neither the mass nor the radius, all bodies of the same shape roll down a given incline with the same acceleration, whatever their mass or size.

Three Remarks From the Textbook

• If the expression for the moment of inertia of a rolling object is of the form $n(MR^2)$, the numerical factor $n$ gives the value of $\dfrac{K^2}{R^2}$ for that object. For a uniform solid sphere $I = \dfrac{2}{5}MR^2 = MK^2$, so $\dfrac{K^2}{R^2} = \dfrac{2}{5}$. Similarly $\dfrac{K^2}{R^2} = 1$ for a ring or a hollow cylinder, $\dfrac{1}{2}$ for a uniform disc or a solid cylinder, and $\dfrac{2}{3}$ for a thin-walled hollow sphere.
• When a rod rolls, it is actually a cylinder that is rolling.
• While rolling, the ratio Translational K.E. : Rotational K.E. : Total K.E. is $1:\dfrac{K^2}{R^2}:\left(1+\dfrac{K^2}{R^2}\right)$. For a hollow sphere, with $\dfrac{K^2}{R^2}=\dfrac{2}{3}$, this is $1:\dfrac{2}{3}:\dfrac{5}{3}$, i.e. $3:2:5$ — so $60\%$ of its kinetic energy is translational and $40\%$ rotational.

Standard Results

• $v = R\omega$ for rolling without slipping; static friction is essential
• $E = \dfrac{1}{2}Mv^2\left[1+\dfrac{K^2}{R^2}\right]$
• $v = \sqrt{\dfrac{2gh}{1+\dfrac{K^2}{R^2}}}$, $\quad a = \dfrac{g\sin\theta}{1+\dfrac{K^2}{R^2}}$, $\quad s = \dfrac{h}{\sin\theta}$
• Pure sliding is the same with $\dfrac{K^2}{R^2}=0$: $v=\sqrt{2gh}$, $a=g\sin\theta$
• Trans. : Rot. : Total K.E. $= 1:\dfrac{K^2}{R^2}:\left(1+\dfrac{K^2}{R^2}\right)$