19.5 Role of friction in rolling  (Page 2/5)

In plain words, it means that if there is an increase in linear velocity, then there shall be an increase in angular velocity as well. So is the correspondence for a decrease in either of two velocities.

We can understand the situation from yet another perspective. Since external force induces linear acceleration, there should be a mechanism to induce angular acceleration so that condition as imposed by the equation of accelerated rolling is met. In other words, the friction appears in magnitude and direction such that above relation is held for rolling.

Static friction in rolling differs to its counterpart in translation in one very important manner. In translation, friction adjusts to the external force parallel to the contact surface completely till the body is initiated. What it means that intermediate static friction is equal to the magnitude of the external force in opposite direction. Such is not the case in rolling i.e. ( $f_S\ne F$ ).

The difference in the nature and magnitude of static friction can be easily understood. In pure translation like in sliding, the sole purpose of friction is to oppose relative motion between surfaces. In the case of rolling, on the other hand, friction coverts a part of one type of acceleration to another (from linear to angular as in this case). This statement may appear a bit awkward. The same can be put more elegantly; if we say that friction changes a part of translational kinetic energy into rotational kinetic energy. This sounds better as we are familiar with the conversion of energy - not conversion of acceleration.

Had it not been friction, then the force passing through COM would have only caused linear acceleration! But, in order to satisfy the physical requirement of rolling as defined by the equation of accelerated rolling - the effect of force is changed from one type to another.

Applying Newton's second law for translation, the linear acceleration of the center of mass is given by :

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

Similarly applying Newton's second law for rotation, the angular acceleration of the center of mass is given by (note that external force causes clockwise rotation and hence negative torque) :

$$\begin{array}{l}\alpha =\frac{\sum \tau }{I}=-\frac{R{f}_S}{I}\end{array}$$

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

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

$$\begin{array}{l}\Rightarrow \frac{F-f_S}{M}=\frac{R^2{f}_S}{I}\end{array}$$

Solving for “ $f_S$ “, we have :

This relation can be used to determine friction in this case. Since all factors like moment of inertia, mass and radius of rotating body are positive scalars, the friction is negative and is in the opposite direction to the applied force.

The relationship as derived above, brings out an interesting feature of rolling friction. Its magnitude depends on the moment of inertia! This is actually expected. The requirement of friction in rolling is for causing angular acceleration, which, in turn, is dependent on moment of inertia. Thus, it is quite obvious that friction (a self adjusting force) should be affected by moment of inertia.