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Q7.32(e):
Read each statement below carefully, and state, with reasons, if it is true or false; (e) A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.

Solution :

Given Variables & Theoretical Foundation

In this problem, we analyze the motion of a rigid body (a wheel) on an inclined plane. We define the following physical parameters:

  • $M$: Mass of the wheel
  • $R$: Radius of the wheel
  • $\theta$: Angle of inclination of the plane
  • $\mu$: Coefficient of friction (Given: $\mu = 0$ for a perfectly frictionless plane)
  • $\tau$: Torque acting on the body
  • $a$: Linear acceleration of the center of mass
  • $\alpha$: Angular acceleration about the center of mass

Step 1: Torque Analysis and Rotational Dynamics

According to Newton's Second Law for rotation, the angular acceleration $\alpha$ is determined by the net external torque acting on the wheel about its center of mass:

$\tau_{ext} = I\alpha$ [Per the Rotational Analog of Newton's Second Law]

On a frictionless surface ($\mu = 0$), the only forces acting on the wheel are:

  • Gravitational force ($Mg$): Acting downwards through the center of mass.
  • Normal force ($N$): Acting perpendicular to the inclined surface.

Because both the weight vector (acting at the center) and the normal force vector (passing through the center) exert zero torque about the center of mass, the net torque is zero ($\tau_{net} = 0$).

$\implies I\alpha = 0 \implies \alpha = 0$ [Since $I \neq 0$]

Step 2: Kinematic Implications of Zero Torque

Since $\alpha = 0$, the wheel does not acquire any angular velocity ($\omega$). If the wheel starts from rest, it will continue to rotate with zero angular velocity throughout its descent. The condition for "rolling without slipping" is defined as:

$v = r\omega$

If $\omega = 0$ while $v > 0$ (due to the component of gravity $Mg \sin\theta$ acting down the plane), then $v \neq r\omega$. This indicates that the motion is purely translational.

Step 3: Defining the Motion State

In physics, "slipping" implies the relative motion between the contact point of the body and the surface. Because there is no friction to provide the necessary torque to initiate rotation, the wheel simply slides down the plane as a particle would. This is classically defined as "pure sliding" or "slipping."

Visual Representation of Forces

Mg N Inclined Plane (Frictionless)

Final Justification

The statement is True. Because the plane is frictionless, no torque is exerted on the wheel. Without torque, the wheel cannot undergo rotational acceleration. Consequently, it translates down the incline without rotating, which constitutes a state of pure slipping.

Final Solution: The statement is True. A wheel on a frictionless inclined plane will undergo translational motion (slipping) because the absence of friction prevents the generation of the torque required for rolling.


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