Rolling Without Slipping On An Inclined Plane

When an object rolls without slipping on an inclined plane, it combines both linear and rotational motion in a fascinating way. This means that the object moves down the slope while spinning around its axis without sliding. Everyday examples of this phenomenon include a rolling can, a ball, or a tyre descending a hill. Understanding this concept helps us appreciate how forces like gravity and friction work together to maintain smooth, energy-efficient motion. In this article, we'll explore the mechanics of rolling without slipping on an inclined plane, examining the forces involved, the equations of motion, and how different shapes and masses affect the rolling behaviour.

This Story also Contains

1. The Linear Velocity of Different Points
2. Rolling Without Slipping on an Inclined Plane
3. Solved Examples Based on Rolling Without Slipping on an Inclined Plane
4. Summary

The Linear Velocity of Different Points

• In pure Translation: All points of a rigid body have the same linear velocity.

• In pure Rotation: All points of a rigid body have the same angular speed but different linear velocities depending on their distance from the axis of rotation.

And in Rolling all points of a rigid body have the same angular speed ($\omega$) but different linear speeds.

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 $K_{\text {net }}=K_T+K_R=\frac{1}{2} m V^2+\frac{1}{2} I \omega^2$

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 the rolling body about its centre ‘O’

And using the Parallel axis theorem

We can write $I_p=I+m R^2$
So we can write $K_{\text {net }}=\frac{1}{2} I_p \omega^2$

Where $I_p$ =moment of inertia of the rolling body about point of contact ‘P’.

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 $\omega$.

Here, the 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 that has zero velocity.)

Net Kinetic Energy for different rolling bodies

${ }_{\mathrm{As}} K_{\text {net }}=K_T+K_R=\frac{1}{2} m V^2\left(1+\frac{K^2}{R^2}\right)$

So the quantity $\frac{K^2}{R^2}$ will have different values for different bodies.

The direction of friction

Kinetic friction will always oppose the rolling motion. While Static friction on the other hand only opposes the tendency of an object to move.

1. When an external force is in the upward diametric part

• - If $K^2=R x$ then no friction will act
  - If $K^2>R x$ then Friction will act in the backward direction
  - If $K^2<R x$ then Friction will act in a forward direction

2. If an external force is in the lower diametric part, Then friction always acts backwards.

Now, let us move to the main topic of our article which is Rolling without Slipping on an Inclined Plane

Rolling Without Slipping on an Inclined Plane

When a body of mass m and radius R rolls down an inclined plane having an angle of inclination ($\theta$) and at height ‘h’

By conservation of mechanical energy

$
m g h=\frac{1}{2} m V^2\left(1+\frac{K^2}{R^2}\right)
$

Where $\mathrm{V}=$ Velocity at the lowest point

And,
$
V=\sqrt{\frac{2 g h}{1+\frac{K^2}{R^2}}}
$

Similarly using $\quad V^2=u^2+2 a s$
$
\text { Acceleration }=a=\frac{g \sin \Theta}{1+\frac{K^2}{R^2}}
$

And angular acceleration $=a=R \alpha$
And we know that $\tau=I \alpha$
And torque due to friction force $=\tau_f=f R=I \alpha=m K^2(R a)$
$f=\frac{m g \sin \Theta}{1+\frac{K^2}{K^2}}$
As $f=\mu N=\mu m g \cos \theta$

So Condition for pure rolling on an inclined plane

$
\mu_s \geq \frac{\tan \Theta}{1+\frac{R^2}{K^2}}
$

Where $\mu_s=$ limiting coefficient of friction
And let $t=$ time taken by the body to reach the lowest point
So using $V=u+a t$

We get,
$
\mathrm{t}=\frac{1}{\sin \theta} \sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}\left(1+\left[\frac{\mathrm{K}^2}{\mathrm{R}^2}\right]\right)}
$

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Solved Examples Based on Rolling Without Slipping on an Inclined Plane

Example 1: A round uniform body of radius $R$ mass $M$ and moment of inertia $I$ rolls down ( without slipping ) an inclined plane making an angle $\Theta$ with the horizontal. Then its acceleration is :

1) $\frac{g \sin \Theta}{1-M R^2 / I}$
2) $\frac{g \sin \Theta}{1+I / M R^2}$
3) $\frac{g \sin \Theta}{1+M R^2 / I}$
4) $\frac{g \sin \Theta}{1-I / M R^2}$

Solution:

Rolling of a body on an inclined plane -

$\begin{aligned}
& a=\frac{g \sin \Theta}{1+\frac{K^2}{R^2}} \\
& f=\frac{m g \sin \Theta}{1+\frac{R^2}{K^2}}
\end{aligned}$

wherein

$\mathrm{K}=$ Radius of gyration
$
\Theta=\text { Angle of inclination }
$

$a=\frac{m g \sin \theta}{m+\frac{l}{R^2}}=\frac{m g \sin \theta}{m\left(1+\frac{l}{m R^2}\right)}=a=\frac{g \sin \theta}{1+\frac{l}{m R^2}}$

Hence, the answer is the option (2).

Example 2: A ring is rolling without slipping on an inclined plane. Find its velocity (in m/sec) after moving on an inclined plane by a distance 10 m, releasing from rest. Assume of inclined of g = 9.8 m/s

1) 5

2) 7

3) 4

4) 3

Solution:

Condition for pure rolling on inclined plane -

$
\mu_s \geq \frac{\tan \Theta}{1+\frac{R^2}{K^2}}
$
wherein
$\mu_s=$ limiting coefficient of friction

By energy conservation

$\begin{aligned} & m g h=\frac{1}{2} m v^2+\frac{1}{2} I W^2 \\ & m g h=\frac{1}{2} m v^2+\frac{1}{2} m K^2\left(\frac{V}{R}\right)^2 \\ & m g h=\frac{1}{2} m v^2\left[1+\frac{K^2}{R^2}\right] \\ & v=\sqrt{\frac{2 g h}{1+\frac{k^2}{R^2}}} \\ & v=\sqrt{\frac{2 \times 9.8 \times 10 \times \sin 30^{\circ}}{1+1}} \Rightarrow v=7 \mathrm{~m} / \mathrm{s}\end{aligned}$

Hence, the answer is option (2).

Example 3: A sphere of radius ' $a$ ' and mass ' $m$ ' rolls along a horizontal plane with constant speed $v_0$, It encounters an inclined plane at angle $\theta$ and climbs upward. Assuming that it rolls without slipping, how far up the sphere will travel ?

1) $\frac{7 v_0^2}{10 g \sin \theta}$
2) $\frac{v_0^2}{5 g \sin \theta}$
3) $\frac{2 v_0^2}{5 g \sin \theta}$
4) $\frac{v_0^2}{2 g \sin \theta}$

Solution:

From energy conservation
$
\begin{aligned}
& \mathrm{mgh}=\frac{1}{2} m v_0^2+\frac{1}{2} I \omega^2 \\
& \mathrm{mgh}=\frac{1}{2} m v_0^2+\frac{1}{2} \times \frac{2}{5} m a^2 \times \frac{v_0^2}{a^2} \\
& g h=\frac{1}{2} v_0^2+\frac{1}{5} v_0^2 \\
& \mathrm{gh}=\frac{7}{10} v_0^2 \\
& \mathrm{~h}=\frac{7}{10} \frac{v_0^2}{g}
\end{aligned}
$
from triangle, $\sin \theta=\frac{h}{l}$
then $\mathrm{h}=l \sin \theta$
$l \sin \theta=\frac{7}{10} \frac{v_0^2}{g}$
$
l=\frac{7}{10} \frac{v_0^2}{g \sin \theta}
$

Hence, the answer is option (1).

Example 4: A solid cylinder of mass m is wrapped with an inextensible light string and, is placed on a rough incline as shown in the figure. The frictional force acting between the cylinder and inclined plane is: (coefficient of friction = 0.4)

1) 5 mg
2) 0
3) $\frac{m g}{5}$
4) $\frac{7}{2} m g$

Let's take solid cylinder is in equilibrium
$
\begin{aligned}
& T+f=m g \sin 60 \\
& T R-f R=0
\end{aligned}
$

Solving we get
$
T=f_{\text {req }}=\frac{m g \sin \theta}{2}
$

But limiting friction $<$ required friction
$
\mu \mathrm{mg} \cos 60^{\circ}<\frac{m g \sin 60^{\circ}}{2}
$
$\therefore$ Hence cylinder will not remain in equilibrium Hence $f=$ kinetic
$
\begin{aligned}
& =\mu_k N \\
& =\mu_k m g \cos 60^{\circ} \\
& =\frac{m g}{5}
\end{aligned}
$

Hence, the answer is option (3).

Example 5: A sphere of mass 2 kg and radius 0.5 m is rolling with an initial speed of 1 $\mathrm{ms}^{-1}$ goes up an inclined plane which makes an angle of $30^{\circ}$ with the horizontal plane without slipping. How long will the sphere take to return to the starting point A ?

1) 0.60 s
2) 0.57 s
3) 0.80 s
4) 0.52 s

Solution:

$\begin{aligned} & \mathrm{a}=\frac{\mathrm{g} \sin \theta}{1+\frac{\mathrm{I}}{\mathrm{mR}^2}}=\frac{5}{7} \times \frac{10}{2}=\frac{25}{7} \\ & \mathrm{t}=\frac{2 \mathrm{v}_0}{\mathrm{a}}=\frac{2 \times 1 \times 7}{25} \\ & =0.56 \mathrm{Sec}\end{aligned}$

Summary

Combining translational and rotational motion results in pure rolling. In a pure rolling state, the body is not affected by the friction force. Consequently, regardless of how rough the surface is, there should ideally be no energy loss via friction. This is a significant rolling property. Since rolling motion loses the least amount of energy, machines with rounded wheels and ball bearings utilise less energy. Pure rolling is a crucial idea in rigid body dynamics with a wide range of applications.