Rotational Dynamics
7.0 Accelerated pure rolling
7.0 Accelerated pure rolling
As discussed in the previous section, pure rolling is a situation when there is no slipping between the two bodies at the point of contact.
For no slipping between the two bodies at the point of contact there should be no relative motion between them.
Therefore, the relative velocity and relative acceleration at the point of contact should be zero.
In the previous section, we discussed condition for pure rolling in three different cases,
7.0.1 Case I
A ball of radius $R$ rolling with velocity $v_b$ and angular velocity $\omega $ on a horizontal ground.
Condition for pure rolling is, $${v_b} = R\omega $$
So, we will differentiate the above equation for the condition of accelerated pure rolling, $$\begin{equation} \begin{aligned} \frac{{d{v_b}}}{{dt}} = R\frac{{d\omega }}{{dt}} \\ {a_b} = R\alpha \\\end{aligned} \end{equation} $$
7.0.2 Case II 
When a ball of radius $R$ rolling with velocity $v_b$ and angular velocity $\omega $ on a plank which is moving with velocity $v_p$ as shown in the figure.
Condition for pure rolling is, $${v_b} - R\omega = {v_p}$$
So, the condition for accelerated pure rolling is, $${a_b} - R\alpha = {a_p}$$
7.0.3 Case III
When a ball is rolling on a inclined wedge with linear velocity $v_b$ and angular velocity $\omega $. The wedge is also moving with horizontal velocity $v_w$ as shown in the figure.
Condition for pure rolling is, $${v_b} = {v_w}\cos \theta + \omega R$$
So, the condition for accelerated pure rolling is, $${a_b} = {a_w}\cos \theta + \alpha R$$
7.0.4 Important steps to solve rotational dynamics problem
| S. No. | Steps |
| 1. | Draw FBD |
| 2. | Apply all the forces |
| 3. | Give the direction of linear acceleration |
| 4. | Give the direction of angular acceleration satisfying the condition of pure rolling |
| 5. | Give the direction of friction. Frictional torque should always support the rotational motion |
| 6. | Solve the equation of translational motion, rotational motion and pure rolling motion |
7.0.5 Force and torque acting on the body for maintaining the accelerated pure rolling
Friction plays an important role in maintaining pure rolling motion.
As we know, friction is self adjusting force. So, it sometimes acts in the forward direction, backward direction or zero depending on the conditions.
Let us consider some examples to understand the concept better.
Illustration 1: Rolling on rough horizontal plane
Consider a rigid body of mass $M$, radius $R$ and moment of inertia $I$ about its axis passing through the centre of mass. The rigid body is kept on a rough ground.
Now a force $F$ is applied at the top most point of the rigid body as shown in the figure.
So, the force acting on the top most point of the rigid body will produce
- linear acceleration $a$ about $COM$
- angular acceleration $\alpha $ about point $O$ i.e. $COM$
As, the rigid body is kept on the rough ground. So, the friction will act on the rigid body at point of contact with the ground i.e. at point $B$.
Let us assume the friction acts in the forward direction as shown in the FBD.
From FBD we can write the following equations.
For translation motion, $$\begin{equation} \begin{aligned} \overrightarrow F + \overrightarrow f = M\overrightarrow a \\ \left( {F + f} \right)\widehat i = Ma\widehat i \\ F + f = Ma\quad ...(i) \\\end{aligned} \end{equation} $$
For rotational motion, $$\begin{equation} \begin{aligned} {\overrightarrow \tau _O} = \left( {{{\overrightarrow R }_{OA}} \times \overrightarrow F } \right) + \left( {{{\overrightarrow R }_{OB}} \times \overrightarrow f } \right) \\ I\overrightarrow \alpha = \left( {{{\overrightarrow R }_{OA}} \times \overrightarrow F } \right) + \left( {{{\overrightarrow R }_{OB}} \times \overrightarrow f } \right)\quad \quad \left[ {As,\overrightarrow \tau = I\overrightarrow \alpha } \right] \\ I\alpha \left( { - \widehat k} \right) = \left( {R\widehat j \times F\widehat i} \right) + \left[ {R\left( { - \widehat j} \right) \times f\widehat i} \right] \\ - I\alpha \widehat k = - FR\widehat k + fR\widehat k \\ I\alpha = (F - f)R \\ F - f = \frac{{I\alpha }}{R}\quad ...(ii) \\\end{aligned} \end{equation} $$
Condition for pure rolling 
$$\begin{equation} \begin{aligned} {\overrightarrow a _{{B_{net}}}} = 0 \\ a\widehat i - R\alpha \widehat i = 0 \\ a = R\alpha \quad ...(iii) \\\end{aligned} \end{equation} $$
From equation $(i)$ & $(iii)$ we get, $$F + f = MR\alpha \quad ...(iv)$$
Now, from equation $(ii)$ & $(iv)$ we get, $$\begin{equation} \begin{aligned} \frac{{F + f}}{{MR}} = \left( {\frac{{F - f}}{I}} \right)R \\ \left( {F + f} \right)I = \left( {F - f} \right)M{R^2} \\ F\left( {I - M{R^2}} \right) = - f\left( {I + M{R^2}} \right) \\ f = \left( {\frac{{M{R^2} - I}}{{M{R^2} + I}}} \right)F \\\end{aligned} \end{equation} $$ or $$\overrightarrow f = \left( {\frac{{M{R^2} - I}}{{M{R^2} + I}}} \right)\overrightarrow F $$
linear acceleration $a$ about $COM$
angular acceleration $\alpha $ about point $O$ i.e. $COM$
| Condition | Result | Diagram |
| $$M{R^2} = I$$ | No friction acts |
|
| $$M{R^2} > I$$ | $\overrightarrow F $ and $\overrightarrow f $ acts in the same direction In the above illustration as $\overrightarrow F $ acts in the forward direction. So, friction $\overrightarrow f $ will also act in the forward direction |
|
| $$M{R^2} < I$$ | $\overrightarrow F $ and $\overrightarrow f $ acts in the opposite direction In the above illustration as $\overrightarrow F $ acts in the forward direction. So, friction $\overrightarrow f $ will act in the backward direction |
|
Illustration 2: Rolling on rough inclined plane
Consider a rigid body of mass $M$, radius $R$ and moment of inertia $I$ rolls down an inclined plane of inclination $\theta $ as shown in the figure.
For translation motion, $$Mg\sin \theta - f = Ma\quad ...(i)$$
For rotational motion (Torque about $COM$), $$\begin{equation} \begin{aligned} {\tau _{COM}} = fR \\ I\alpha = IR\quad ...(ii) \\\end{aligned} \end{equation} $$
For pure rolling motion, $$a = R\alpha \quad ...(iii)$$
From equation $(i)$, $(ii)$ and $(iii)$ we get, $$\begin{equation} \begin{aligned} Mg\sin \theta - f = MR\alpha \quad {\text{or}}\quad \alpha = \frac{{Mg\sin \theta - f}}{{MR}} \\ fR = I\alpha \quad {\text{or}}\quad \alpha = \frac{{fR}}{I} \\\end{aligned} \end{equation} $$ So, $$\begin{equation} \begin{aligned} \frac{{fR}}{I} = \frac{{Mg\sin \theta - f}}{{MR}} \\ \\ f = \frac{{Mg\sin \theta }}{{1 + \frac{{M{R^2}}}{I}}}\quad ...(iv) \\\end{aligned} \end{equation} $$
So, we can see whatever be the value of $I$ friction will always acts in the backward direction.
From equation $(ii)$, $(iii)$ and $(iv)$ we get, $$\begin{equation} \begin{aligned} \left( {\frac{{Mg\sin \theta }}{{1 + \frac{{M{R^2}}}{I}}}} \right)R = \frac{{Ia}}{R} \\ \\ a = \frac{{M{R^2}g\sin \theta }}{{\left( {1 + \frac{{M{R^2}}}{I}} \right)I}} \\ \\ a = \frac{{g\sin \theta }}{{1 + \frac{I}{{M{R^2}}}}} \\\end{aligned} \end{equation} $$
Rigid body having greater $I$ will have less acceleration. So, it takes more time to reach down.
For pure rolling to takes place, friction should be less than the maximum friction.
As we know, $$\begin{equation} \begin{aligned} N = Mg\cos \theta \\ {f_{\max }} = \mu N = \mu Mg\cos \theta \\ f \leqslant {f_{\max }} \\\end{aligned} \end{equation} $$ So, $$\begin{equation} \begin{aligned} \frac{{Mg\sin \theta }}{{1 + \frac{{M{R^2}}}{I}}} = \mu Mg\cos \theta \\ \\ \mu \geqslant \frac{{\tan \theta }}{{1 + \frac{{M{R^2}}}{I}}} \\\end{aligned} \end{equation} $$
where $\theta $ is the inclination of the wedge
Illustration 3: When force $F$ acts at a distance $r$ from the centre of mass
Consider a rigid body of mass $M$, radius $R$ and moment of inertia $I$. A force $F$ is applied at a distance $r$ from the centre of mass as shown in the figure.
From FBD we can write, $$\begin{equation} \begin{aligned} \overrightarrow \alpha = \alpha \left( { - \widehat k} \right) \\ \overrightarrow a = a\left( {\widehat i} \right) \\ \overrightarrow F = F\left( {\widehat i} \right) \\ \overrightarrow f = f\left( { - \widehat i} \right) \\\end{aligned} \end{equation} $$
For rotational motion (Torque about $COM$),$$\begin{equation} \begin{aligned} {\overrightarrow \tau _{net}} = \left( {\overrightarrow R \times \overrightarrow f } \right) + \left( {\overrightarrow r \times \overrightarrow F } \right) \\ I\overrightarrow \alpha = \left( {\overrightarrow R \times \overrightarrow f } \right) + \left( {\overrightarrow r \times \overrightarrow F } \right) \\ I\alpha \left( { - \widehat k} \right) = \left[ {R\left( { - \widehat j} \right) \times f\left( { - \widehat i} \right)} \right] + \left[ {r\left( { - \widehat j} \right) \times F\left( {\widehat i} \right)} \right] \\ - I\alpha \widehat k = - fR\widehat k + Fr\widehat k \\ I\alpha = fR - Fr\quad ...(i) \\\end{aligned} \end{equation} $$
For translational motion, $$\begin{equation} \begin{aligned} F\left( {\widehat i} \right) - f\left( {\widehat i} \right) = Ma\left( {\widehat i} \right) \\ F - f = Ma\quad ...(ii) \\\end{aligned} \end{equation} $$
For pure rolling condition, $$a = R\alpha \quad ...(iii)$$
From equation $(i)$, $(ii)$ and $(iii)$ we get, $$\begin{equation} \begin{aligned} (fR - Fr) = I\alpha \quad {\text{or}}\quad \alpha = \frac{{\left( {fR - Fr} \right)}}{I} \\ (F - f) = MR\alpha \quad {\text{or}}\quad \alpha = \frac{{(F - f)}}{{MR}} \\\end{aligned} \end{equation} $$ So, $$\begin{equation} \begin{aligned} \frac{{(F - f)}}{{MR}} = \frac{{\left( {fR - Fr} \right)}}{I} \\ \\ f = \left( {\frac{{MRr + I}}{{M{R^2} + I}}} \right)F \\\end{aligned} \end{equation} $$ or $$\overrightarrow f = - \left( {\frac{{MRr + I}}{{M{R^2} + I}}} \right)\overrightarrow F $$
Note: -ve sign is because forces $\overrightarrow f $ and $\overrightarrow F $ are in opposite direction.
