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22 Questions around this concept.
A string is wound around a hollow cylinder of mass 5 kg and radius 0.5m. If the string is now pulled with a horizontal force of 40N, and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration ( in rad/s2 ) of the cylinder will be (neglect the mass and thickness of the string):
A solid sphere and cylinder of identical radii and mass approach an incline with the same linear velocity (see figure ). Both roll without slipping all throughout. The two climb maximum heights $h_{s p h}$ and $h_{c y l}$ on the incline. The ratio $\frac{h_{s p h}}{h_{c y l}}$ is given by:

Two forces of magnitude F and 2F are acting on a uniform disc kept on a horizontal rough surface as shown in Fig. Friction force by the horizontal surface on the also:

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A uniform solid cylindrical roller of mass'm' is being pulled on a horizontal surface with force F parallel to the surface and applied at its center. If the acceleration of the cylinder is 'a' and it is rolling without slipping, then the value of 'F' is:
A ring of radius R is rolling over a rough horizontal surface with velocity $v_0$. Two points located at A and B on rim of ring. The angular velocity of point A w.r.t. point B will be

A system of uniform cylinders and plates is shown in figure. All the cylinders are identical and there is no slipping at any contact. Velocity of lower and upper plate is and
respectively as shown in figure. The ratio of angular speed of the upper cylinders to lower cylinders is:-
A tangential force F acts at the top of a thin spherical shell of mass m and radius R . The acceleration of the shell if it rolls without slipping -
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A wheel is doing pure rolling motion when a force F acts on the centre of the wheel. The linear acceleration of the wheel (centre). When the mass of the wheel is m kg this mass can be assumed to be concentrated at the rim having radius $R$.
A ring of mass $\mathrm{m}$ and radius r rolls down (without slipping) down an inclined plane starting from rest. If the centre of the ring is at a height $\mathrm{h}$, the angular velocity when the ring reaches the base is -
A uniform disc of mass and radius
rolls, without slipping down a fixed plane inclined at an angle
to the horizontal. The linear acceleration of the disc is:
Linear velocity of different points
In pure Translation-

In pure Rotation-

And in Rolling all points of a rigid body have same angular speed () but different linear speed.
I.e 
During Rolling motion
$\begin{aligned} & \text { If } V_{c m}>R w \rightarrow \text { slipping motion } \\ & { }_{\text {If }} V_{c m}=R w \rightarrow \text { pure rolling } \\ & { }_{\text {If }} V_{c m}<R w \rightarrow \text { skidding motion }\end{aligned}$
When the object rolls across a surface such that there is no relative motion of the object and surface at the point of contact, the motion is called rolling without slipping.

Here the point of contact is P.
Friction force is available between object and surface but work done by it is zero because there is no relative motion between body and surface at the point of contact.
Or we can say No dissipation of energy is there due to friction.
I.e., Energy is conserved.
Which is $K_{n e t}=K_T+K_R=\frac{1}{2} m V^2+\frac{1}{2} I \omega^2$
Now using $V=\omega \cdot R$
And using $K_{\text {net }}=\frac{1}{2} m V^2+\frac{1}{2} I \omega^2=\frac{1}{2}\left(I+m R^2\right) \omega^2$
Where I = moment of inertia of rolling body about its center ' $O$ '
And using the Parallel axis theorem
We can write $I_p=I+m R^2$
So we can write $K_{n e t}=\frac{1}{2} I_p \omega^2$
Where $I_{p=\text { moment of inertia of a rolling body about point of contact ' } P \text { '. }}$
So this Rolling motion of a body is equivalent to a pure rotation about an axis passing through the point of contact (here through P) with the same angular velocity .
Here axis passing through the point of contact P is also known as the Instantaneous axis of rotation.
(Instantaneous axis of rotation-Motion of an object may look as pure rotation about a point which has zero velocity.)
Net Kinetic Energy for different rolling bodies
$$
{ }_{\mathrm{As}} K_{n e t}=K_T+K_R=\frac{1}{2} m V^2\left(1+\frac{K^2}{R^2}\right)
$$
So the quantity $\frac{K^2}{R^2}$ will have different values for different bodies.
|
Rolling body |
$\frac{K^2}{R^2}$ | $K_{n e t}$ |
|
Ring Or Cylindrical shell |
1 |
$m V^2$ |
|
Disc Or solid cylinder |
$\frac{1}{2}$ | $\frac{3}{4} m V^2$ |
|
Solid sphere |
$\frac{2}{5}$ | $\frac{7}{10} m V^2$ |
|
Hollow sphere |
$\frac{2}{3}$ | $\frac{5}{6} m V^2$ |
Direction of fiction-
Kinetic friction will always oppose the rolling motion. While Static friction on the other hand only opposes the tendency of an object to move.
When an external force is in an upward diametric part
1. If $K^2=R x$ then-No friction will act
2. If $K^2>R x$ then Friction will act in backward direction
3. If $K^2<R x$ then Friction will act in a forward direction
If external force is in the lower diametric part,
Then friction always acts backward
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