Rolling Motion

Rolling as translation plus rotation

NCERT opens the chapter by contrasting two motions on the same incline. A block sliding down the slope is in pure translation: every particle of the block has the same velocity at any instant. A cylinder rolling down the same slope shifts from top to bottom, so it too has translational motion — but, as NCERT states, "all its particles are not moving with the same velocity at any instant." The cylinder is therefore not in pure translation. Its motion is "translational plus something else", and that something else is rotational motion.

So rolling is the superposition of two motions: the centre of mass translates forward with velocity $v_{cm}$, and the body simultaneously rotates about the axis through the centre with angular speed $\omega$. Every point on the rim carries both contributions at once. This single idea — add the translation of the centre to the rotation about the centre — generates every quantitative result on this page.

The rolling-without-slipping condition

The defining feature of rolling without slipping is that the body does not skid against the surface. NCERT records the consequence directly in the caption of Fig. 6.2: "the velocity of the point of contact $P_3$ is zero at any instant, if the cylinder rolls without slipping."

For that to happen, the forward rim speed produced by rotation, $\omega R$, must exactly equal the forward translational speed of the centre, $v_{cm}$. This is the kinematic constraint that NCERT and NIOS both use for pure rolling:

$$ v_{cm} = \omega R $$

Differentiating both sides with respect to time, and noting $R$ is constant, gives the matching condition on accelerations:

$$ a_{cm} = \alpha R $$

where $a_{cm}$ is the linear acceleration of the centre and $\alpha$ the angular acceleration. Both conditions must hold at every instant for the motion to stay pure rolling. If $v_{cm} > \omega R$ the body skids forward (translation outruns rotation); if $v_{cm} < \omega R$ it spins in place like a stuck wheel. Only $v_{cm}=\omega R$ is true rolling.

Velocity of the contact point and the top

Treat the velocity of any point on the rim as the vector sum of two pieces — the translation $v_{cm}$ of the centre (forward, the same for every point) and the rotational velocity $\omega R$ tangent to the rim at that point. At the three special points this superposition gives clean numbers.

• Contact point (bottom): rotation drives the rim backward at $\omega R$ while translation carries it forward at $v_{cm}$. Since $v_{cm}=\omega R$, they cancel: $v_{contact} = v_{cm} - \omega R = 0$.
• Centre: there is no rotational contribution at the axis itself, so $v_{centre} = v_{cm}$.
• Top point: rotation drives the rim forward at $\omega R$ in the same direction as translation, so $v_{top} = v_{cm} + \omega R = 2v_{cm}$.

Because the contact point is momentarily at rest, the whole wheel can be regarded at that instant as rotating purely about the contact point. This is why the top moves at twice the centre's speed — it is twice as far from the instantaneous axis as the centre is.

Kinetic energy of a rolling body

The total kinetic energy of a rolling body is the sum of its translational and rotational kinetic energies. NIOS §7.5 writes them as $(KE)_{tr} = \tfrac{1}{2}Mv_{cm}^2$ and $(KE)_{rot} = \tfrac{1}{2}I\omega^2$, with the total being their sum:

$$ KE = \tfrac{1}{2}m\,v_{cm}^2 + \tfrac{1}{2}I_{cm}\,\omega^2 $$

Now introduce the radius of gyration $k$, defined by $I_{cm} = mk^2$, and use the rolling condition $\omega = v_{cm}/R$. Substituting:

$$ KE = \tfrac{1}{2}m v_{cm}^2 + \tfrac{1}{2}(mk^2)\!\left(\frac{v_{cm}}{R}\right)^{\!2} = \tfrac{1}{2}m v_{cm}^2\!\left(1 + \frac{k^2}{R^2}\right) $$

The bracket $\left(1 + \dfrac{k^2}{R^2}\right)$ is the master factor of rolling. The "1" is the translational share; the "$k^2/R^2$" is the rotational share. The fraction of energy that is translational is therefore

$$ \frac{KE_{tr}}{KE} = \frac{1}{\,1 + k^2/R^2\,} $$

For a solid sphere $k^2/R^2 = 2/5$, so this fraction is $\dfrac{1}{1+2/5} = \dfrac{5}{7}$ — exactly the NEET 2018 result. The more a body's mass sits near its rim (larger $k^2/R^2$), the larger the rotational share of its energy.

The role of friction in rolling

Friction plays a paradoxical double role in rolling. On one hand, friction at the contact is what makes rolling possible at all: it supplies the torque that spins the body up so that $\omega R$ keeps pace with $v_{cm}$. Without friction, a body placed on an incline would simply slide, not roll. On the other hand, in pure rolling this friction is static friction, and the point at which it acts — the contact point — is instantaneously at rest.

A force does work only when its point of application moves. Since the contact point has zero velocity, the static friction there does no work. This is the deep reason NIOS can apply straightforward energy conservation to a body rolling down an incline: "the loss of energy due to friction is small and can be neglected" in ideal rolling, so the potential energy at the top equals the kinetic energy at the bottom.

Rolling down an incline

Consider a body of mass $M$, radius $R$ and moment of inertia $I = Mk^2$ released from rest at height $h$ on an incline of angle $\theta$. NIOS Example 7.8 applies energy conservation: the potential energy at the top equals the total kinetic energy at the bottom,

$$ Mgh = \tfrac{1}{2}Mv^2 + \tfrac{1}{2}I\omega^2 . $$

Using $v=\omega R$ and $I = Mk^2$, the right side becomes $\tfrac{1}{2}Mv^2(1+k^2/R^2)$, so the speed at the bottom is

$$ v = \sqrt{\dfrac{2gh}{\,1 + k^2/R^2\,}} . $$

This speed depends only on $h$ and the shape factor $k^2/R^2$ — mass and radius cancel completely. NIOS makes the point for a hoop: its speed at the bottom is "independent of mass and radius", so "a hoop of any material and any radius rolls down with the same speed on the inclined plane."

The acceleration of the centre down the slope follows from the same physics. The component of gravity along the incline is $Mg\sin\theta$, but only a fraction of it goes into linear acceleration because the rest must spin the body up. The standard NCERT-derived result, which the PYQ solutions on this page use directly, is

$$ a = \dfrac{g\sin\theta}{\,1 + k^2/R^2\,} . $$

Again, mass and radius are absent. The single quantity that controls both the speed at the bottom and the acceleration down the slope is the dimensionless shape factor $k^2/R^2$.

The race — ring vs disc vs sphere

Release a ring, a disc and a solid sphere from rest at the top of the same incline. Which reaches the bottom first? Since $a = g\sin\theta/(1+k^2/R^2)$, the body with the smallest $k^2/R^2$ has the largest acceleration and wins — regardless of mass or radius. The factor grid below collects the standard values.

So the finishing order down the incline is fixed: solid sphere first, disc second, ring last. The sphere spends the smallest fraction of its energy on rotation, leaving the most for getting to the bottom; the ring, with all its mass at the rim, spends the most on rotation and trails. This is precisely the answer to NEET 2016 Q.167, where a disk and a sphere of the same radius but different masses race — the sphere wins, mass irrelevant.

Worked examples