Rolling Motion

Rolling motion is a type of movement where an object, such as a wheel or a ball, rotates about an axis while simultaneously moving forward along a surface without slipping. This combined rotational and translational motion allows the object to cover distance efficiently. In rolling motion, every point on the object follows a circular path around the axis of rotation, and the point in contact with the surface momentarily remains at rest relative to the surface. This motion is fundamental in many mechanical systems and everyday activities, enabling smooth and controlled movement.

1.0Rolling Motion

• When a body performs translatory motion as well as rotatory motion combinedly then it is said to undergo rolling motion. The velocity of the centre of mass represents linear motion, while the angular velocity about the centroid axis represents rotatory motion.

• In pure translational motion, all points move with the same linear velocity V_C. In pure rotational motion about COM, all points move with the same angular velocity about the central axis.

Velocity of Particle on a Rolling Object:

• Velocity of particles on a rolling object can be seen by considering the translatory motion of the body and rotational motion of the body wrt an axis passing through the centre of mass.

2.0Rolling Without Slipping (Pure Rolling)

• If the relative velocity of the point of contact of the rolling body with the surface is zero then it is known as pure rolling.
• If a body is performing rolling then the velocity of any point of the body with respect to the surface is given by $\vec{v}={\vec{v}}_{CM}+\vec{\omega }\times \vec{R}$

$\begin{array}{rl}& v_A=v_{CM}+\omega R=v_{CM}+v_{CM}=2v_{CM}\\ & v_B=v_{CM}+\frac{\omega R}{2}=v_{CM}+\frac{v_{CM}}{2}=\frac{3}{2}{v}_{CM}\\ & v_C=v_{CM}+0=v_{CM}\\ & v_D=v_{CM}-\frac{\omega R}{4}=v_{CM}-\frac{v_{CM}}{4}=\frac{3}{4}{v}_{CM}\\ & v_E=v_{CM}-\omega R=v_{CM}-v_{CM}=0\end{array}$

3.0For Contact Point on Ground

[Condition for Pure Rolling]

Case-1-If $v_{CM}<\omega R$

Point (will slip backward wrt ground. It is called rolling with slipping.

Case-2-If $v_{CM}>\omega R$

Point C will slip forward with respect to ground. It is also called rolling with slipping

Case-3-If $v_{CM}=\omega R$

The velocity of point C will be zero. So it will not slip with respect to ground. It is called pure rolling. So, for pure rolling on ground.

$v_{CM}=\omega R$

For pure rolling motion of the above mentioned body, we have

$\begin{array}{rl}& v_A=2V_{CM}\\ & v_E=\sqrt{2}{V}_{CM}\\ & v_F=\sqrt{2}{V}_{CM}\\ & v_B=0\end{array}$

4.0Rolling With Slipping

• When a body rolls on a surface under external force, the frictional force on the body (if any) will be static in nature, less than its limiting value. But if the object rolls with slipping the nature of friction should be kinetic in nature.

Nature of friction in the following cases assuming the body is perfectly rigid

1. Condition for Pure Rolling

Contact point A on the object does not slip relative to point B on the surface. So velocity of point A must be equal to velocity of point B.

$\begin{array}{rl}& V_A=V_B\\ & V_c-\omega R=V_{\text{surface }}\end{array}$

No, friction and pure rolling.

2. $V_c=\omega R$

Rigid, then there is a small friction acting in this case is called rolling friction.

3. $V_c>\omega R$ or $V_c<\omega R$

No friction force but not pure rolling.

4. $V_c>\omega R$

If $V_c>\omega R$ contact point slips forward. So kinetic friction will act in backward direction, so Vc decreases and $\omega$ increases.

Ultimately,when $V_c=\omega R$, pure rolling starts

5. $V_c<\omega R$

If $V_c<\omega R$,contact point slips in backward direction, so kinetic friction will act in forward direction, this kinetic friction will act in forward direction, this kinetic friction will increase V and $\omega$ decrease, so after some time.Ultimately, $V_c=\omega R$ , pure rolling will starts.

6. $V_c=\omega R$ (Initial)

No friction and no pure rolling

7. $V_c=\omega R$ (Initial)

Static friction, whose value can lie between zero and μN will act in backward direction. If co-efficient of friction is sufficiently high, then f_s compensates for increasing V due to F by increasing $\omega$ and body may continue in pure rolling with increasing V as well as $\omega$.

5.0Accelerated Pure Rolling

For Pure Rolling $V_c=\omega R$

To maintain pure rolling in acceleration motion, if $V_c$ increases, so $\omega$ decreases.

$\frac{d{V}_c}{dt}=\frac{d}{dt}(\omega R)=\frac{Rd\omega }{dt}$

$\Rightarrow a_{COM}=\alpha R\to$Condition for accelerated pure rolling

$\alpha$=Angular Acceleration

$a_{COM}$ = acceleration of COM

$\begin{array}{rl}& {\left({\vec{a}}_{\text{net }}\right)}_P={\vec{a}}_{\text{translation }}+{\vec{a}}_{\text{rotation }}\\ & {\left({\vec{a}}_{\text{net }}\right)}_P={\vec{a}}_{\text{COM }}+{\vec{a}}_{\text{centripetal }}+{\vec{a}}_{\text{tangential }}\end{array}$

If Surface is Also Moving

For Pure Rolling

$\begin{array}{rl}& V_A=V_B\\ & V_c-\omega R=V_{\text{surface }}\end{array}$

And

$\begin{array}{rl}& a_{T_A}=a_{T_B}\text{ i.e. }\\ & \therefore a_{\text{COM }}-\alpha R=a_{\text{plank }}\end{array}$

6.0Rolling Motion on An Inclined Plane

A body of mass M and radius R is rolling down a plane inclined at an angle with the horizontal. The body rolls without slipping. The centre of mass of the body moves in a straight line. External forces acting on the body are :  

• Weight Mg of the body vertically downwards through its center of mass.
• The normal reaction N of the inclined plane.
• The frictional force f acting upwards and parallel to the inclined plane.

For Linear Motion $Mg\sin \theta -f=M{a}_{CM}$

For angular motion $\tau =fR=I\alpha$

From the condition for pure rolling 

$a_{CM}=\alpha R\Rightarrow Mg\sin \theta -f=M\left(\frac{f{R}^2}{I}\right)=M\left(\frac{f{R}^2}{M{K}^2}\right)\Rightarrow f=\frac{Mg\sin \theta }{\left(1+\frac{R^2}{K^2}\right)}$

But $f\le \mu Mg\cos \theta \Rightarrow \frac{Mg\sin \theta }{\left(1+\frac{k^2}{k^2}\right)}\le \mu Mg\cos \theta$

$\Rightarrow \mu \ge \frac{\tan \theta }{\left(1+\frac{k^2}{k^2}\right)}\Rightarrow \text{ For Pure Rolling }{\mu }_{\min }=\frac{\tan \theta }{\left(1+\frac{k^2}{k^2}\right)}$

7.0Rolling Motion on An Inclined Plane

1. Velocity at the bottom of the inclined plane

Applying Conservation of mechanical energy principle

$mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

$mgh=\frac{1}{2}m{v}^2+\frac{1}{2}m{K}^2\left(\frac{v^2}{R^2}\right)$

$mgh=\frac{1}{2}m{v}^2\left(1+\frac{K^2}{R^2}\right).........(1)$

$h=s\sin \theta \dots \dots (2)$

from(1) and (2)

$v_{\text{Rolling }}=\sqrt{\frac{2gh}{1+\frac{K^2}{R^2}}}=\sqrt{\frac{2gs\sin \theta }{1+\frac{K^2}{R^2}}}$

When the body slides $\frac{K^2}{R^2}=0$

Velocity when the body slides

$v_{\text{sliding }}=\sqrt{2gh}=\sqrt{2gs\sin \theta }\Rightarrow v_{\text{sliding }}>v_{\text{Rolling }}$

2. Acceleration of the body

$\begin{array}{rl}& v^2=u^2+2as(\because u=0)\\ & \therefore v^2=2as\\ & \frac{2gs\sin \theta }{\left(1+\frac{K^2}{R^2}\right)}=2as\Rightarrow a_{\text{rolling }}=\frac{g\sin \theta }{\left(1+\frac{K^2}{R^2}\right)}\end{array}$

When the body slides $\frac{K^2}{R^2}=0$

Acceleration when the body slides $a_{\text{sliding }}=g\sin \theta \Rightarrow a_{\text{sliding }}>a_{\text{rolling }}$

3. Time taken by the rolling body to reach the bottom

$s=\frac{1}{2}a{t}^2\Rightarrow t=\sqrt{\frac{2s}{a}},s=\frac{h}{\sin \theta }$

$t_{\text{Rolling }}=\sqrt{\frac{2s}{g\sin \theta }\left(1+\frac{K^2}{R^2}\right)}=\sqrt{\frac{2h}{g{\sin }^2\theta }\left(1+\frac{K^2}{R^2}\right)}=\frac{1}{\sin \theta }\sqrt{\frac{2h}{g}\left(1+\frac{K^2}{R^2}\right)}$

When the body slides $\frac{K^2}{R^2}=0$

Time of descent when the body slides $t_{\text{sliding }}=\sqrt{\frac{2s}{g\sin \theta }}=\sqrt{\frac{2h}{g{\sin }^2\theta }}=\frac{1}{\sin \theta }\sqrt{\frac{2h}{g}}$

So, $t_{\text{rolling }}>t_{\text{sliding }}$,

Note: If different bodies are allowed to roll down an inclined plane, then the body with

1. Least $\frac{K^2}{R^2}$will reach first
2. Maximum $\frac{K^2}{R^2}$ will reach last
3. Equal $\frac{K^2}{R^2}$ will reach together
4. Change in kinetic energy due to rolling $v_2>v_1$

$\frac{1}{2}m{v}_2^2\left(1+\frac{K^2}{R^2}\right)-\frac{1}{2}m{v}_1^2\left(1+\frac{K^2}{R^2}\right)=\frac{1}{2}m\left(1+\frac{K^2}{R^2}\right)\left(v_2^2-v_1^2\right)$

5. Rolling v/s Sliding

$v_{\text{rolling }}=\frac{v_{\text{sliding }}}{\sqrt{1+\frac{K^2}{R^2}}},a_{\text{rolling }}=\frac{a_{\text{sliding }}}{1+\frac{K^2}{R^2}},t_{\text{rolling }}=t_{\text{sliding }}\sqrt{1+\frac{K^2}{R^2}}$

$v_{\text{rolling }}<v_{\text{sliding }},a_{\text{rolling }}<a_{\text{sliding }},t_{\text{rolling }}<t_{\text{sliding }}$

Illustration-1 If pure rolling occurs find relation between $\alpha$ and a ?

Solution: Condition

$a_{T_1}=a_2$

$\begin{array}{rl}& a-R\alpha =a_2\left(\because \text{ the surface is fixed, }\therefore a_2=0\right)\\ & a=R\alpha \end{array}$

Illustration-2. A body of mass M and radius r, rolling with velocity v on a smooth horizontal floor, rolls up a rough irregular inclined plane up to a vertical height$\frac{3v^2}{4g}$. Compute the moment of inertia of the body and comment on its shape?

Solution: 

$K{E}_{\text{Total }}=K{E}_{Tr}+K{E}_{\text{Rot }}=\frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2$

$E=\frac{1}{2}M{v}^2\left[1+\left(\frac{I}{M{r}^2}\right)\right][asv=r\omega ]$

When it rolls up on an irregular inclined plane of height $\frac{3v^2}{4g}$., its KE is fully converted into PE. So, by conservation of mechanical energy.

$\frac{1}{2}M{v}^2\left[1+\frac{1}{M{r}^2}\right]=Mg\left[\frac{3v^2}{4g}\right]$

which, on simplification, gives.