19.5 Role of friction in rolling  (Page 3/5)

The line of action of external force does not pass through center of mass.

Now, we consider the second case. For the sake of contrast, we consider a tangential force acting tangentially at the top as shown in the figure below. The external force that does not pass through center of mass causes angular acceleration apart from causing linear acceleration. The force has dual role to play in this case. As far as its rotational form (torque) is concerned, resulting angular acceleration means that angular velocity tends to increase. This, in turn, induces tendency of the rolling body to slide in the backward direction (i.e. in the opposite direction of force). Force of friction, therefore, appears in the direction of external force to check the sliding tendency.

The friction acts to balance the changes in a manner so that condition of rolling is met. First, it enhances the net external force ( $F+f_S$ ) and hence the translational acceleration( $a_C$ ). Second, it constitutes a torque in anticlockwise direction inducing angular deceleration. In the nutshell, an increase in angular acceleration due to net torque is moderated by friction by a two pronged actions and the rolling is maintained even when the body is accelerated in rotation.

We can understand the situation from yet another perspective. Since external torque induces angular acceleration, there should be a mechanism to induce linear acceleration so that condition of accelerated rolling is met.

Friction appears in magnitude and direction such that above relation is held for rolling. The linear acceleration of the center of mass is given by :

$$\begin{array}{l}{a}_C=\frac{\sum F}{M}=\frac{F+f_S}{M}\end{array}$$

The angular acceleration of the center of mass is given by (note that external force causes clockwise rotation and hence negative torque) :

$$\begin{array}{l}\alpha =\frac{\sum \tau }{I}=\frac{R(f_S-F)}{I}\end{array}$$

The two accelerations are such that they are related by the equation of accelerated rolling (negative sign as linear and angular accelerations are in opposite directions) as :

$$\begin{array}{l}{a}_C=-\alpha R\end{array}$$

$$\begin{array}{l}\Rightarrow \frac{F+f_S}{M}=\frac{R(f_S-F)}{I}\end{array}$$

Solving for “ $f_S$ “, we have :

This relation (note that this is not a general case, but a specific case of force acting tangentially at the top of the rolling body) provides a reflection on the following aspects of friction force :

• Since $I>0\mathrm{and}I<M{R}^2$ , friction force, “ $f_S$ ”, is positive and is in the direction of external force. We should note here that for all rigid body $M{R}^2$ represents the maximum moment of inertia which corresponds to a ring or hollow cylinder. MIs of all other bodies like sphere, disk etc. are less than this maximum value.
• For ring and hollow cylinder, $I=M{R}^2$ . Thus, friction is zero even for accelerated rolling in the case of these two rigid body. This is one of the reasons that wheels are made to carry more mass on the circumference.

We conclude from the discussion as above that friction plays the role of maintaining the rolling motion in acceleration. As a matter of fact, had it not been the friction, it would have been possible to have accelerated rolling – it would not have been possible to accelerate bicycle, car, motor, rail etc! Can you imagine these vehicles moving with a constant velocity! There would not have been any car racing either without friction! Such is the role of friction in rolling.