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Q7.29(b):

Explain why friction is necessary to make the disc in Fig. 7.41 roll in the direction indicated. (b) What is the force of friction after perfect rolling begins ?

Solution :

1. Role of Friction in Rolling Motion

To initiate rolling motion of a disc on a horizontal surface, an external torque must be applied to overcome inertia. Consider a disc of mass $M$ and radius $R$. When a force $F$ is applied at the center, the disc tends to slide forward. [Per Newton's First Law], an object in motion stays in motion; however, to induce rotation, an angular acceleration $\alpha$ is required.

  • Translational motion: The force $F$ causes a linear acceleration $a = F/M$.
  • Rotational motion: For the disc to roll, there must be a torque $\tau$ about the center of mass, where $\tau = I\alpha$.
  • Mechanism: If there were no friction, the disc would merely translate without rotation. Friction $f$ acts at the contact point in the direction opposite to the potential sliding motion. This creates a torque $\tau = f \cdot R$ which provides the necessary angular acceleration $\alpha = \frac{fR}{I}$.

2. Force of Friction in Perfect Rolling

Perfect rolling (rolling without slipping) is defined by the condition $v = \omega R$ and $a = \alpha R$. Let us derive the force of friction $f$ for a disc ($I = \frac{1}{2}MR^2$) being pushed by a force $F$ at its center.

Step 1: Write the equations of motion.

Translational: $F - f = Ma$ [By Newton's Second Law]

Rotational: $fR = I\alpha = \left(\frac{1}{2}MR^2\right) \left(\frac{a}{R}\right)$ [By $\tau = I\alpha$]

Step 2: Simplify the rotational equation.

$fR = \frac{1}{2}MRa \implies f = \frac{1}{2}Ma$

Step 3: Substitute into the translational equation.

$F - \frac{1}{2}Ma = Ma$

$F = \frac{3}{2}Ma \implies a = \frac{2F}{3M}$

Step 4: Solve for friction $f$.

$f = \frac{1}{2}M \left(\frac{2F}{3M}\right) = \frac{1}{3}F$

3. Visual Representation of Forces

F f Contact Point

Analytical Summary

Parameter Expression
Linear Acceleration ($a$) $\frac{2F}{3M}$
Angular Acceleration ($\alpha$) $\frac{2F}{3MR}$
Frictional Force ($f$) $\frac{F}{3}$

Once perfect rolling is established at a constant velocity, the acceleration $a$ becomes zero if the external driving force $F$ is removed. In the absence of an external force, $f = 0$. However, as long as the accelerating force $F$ is applied to the center of the disc, the frictional force required to maintain the rolling condition is exactly one-third of the applied force.

Final Solution: The force of friction after perfect rolling begins (assuming the force $F$ is maintained) is $f = \frac{1}{3}F$.


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