Question

Consider a uniform solid disk of mass \(M\) and radius \(R\), rolling without slipping down an incline which is at angle \(\gamma\) to the horizontal. The instantaneous point of contact between the disk and the incline is called \(P\). (a) Draw a free-body diagram, showing all forces on the disk. (b) Find the linear acceleration \(\dot{v}\) of the disk by applying the result \(\dot{\mathbf{L}}=\Gamma^{\text {ext }}\) for rotation about \(P .\) (Remember that \(L=I \omega\) and the moment of inertia for rotation about a point on the circumference is \(\frac{3}{2} M R^{2} .\) The condition that the disk not slip is that \(v=R \omega \text { and hence } \dot{v}=R \dot{\omega} .)\) (c) Derive the same result by applying \(\dot{\mathbf{L}}=\mathbf{\Gamma}^{\mathrm{ext}}\) to the rotation about the CM. (In this case you will find there is an extra unknown, the force of friction. You can eliminate this by applying Newton's second law to the motion of the CM. The moment of inertia for rotation about the \(\mathrm{CM}\) is \(\frac{1}{2} M R^{2}\).)

Short answer

The linear acceleration \( \dot{v} \) is \( \frac{2}{3}g\sin\gamma \).

Step-by-step solution

Sketch the Free-Body Diagram

Draw the disk on an inclined plane. Identify the forces acting on the disk: gravitational force \(Mg\) acting downward through the center of mass, normal force \(N\) perpendicular to the incline, and frictional force \(f\) acting up the incline at point \(P\).

Apply Rotational Dynamics About Point P

Use the equation \(\dot{\mathbf{L}}=\Gamma^{\text{ext}}\). For rotation about \(P\), calculate the torque due to gravity as \(MgR\sin\gamma\) since friction and normal force pass through \(P\) and produce no torque. Set \(\dot{L} = I_P \dot{\omega} = \frac{3}{2}MR^2 \frac{\dot{v}}{R}\) equal to this torque, giving \[\frac{3}{2}MR^2 \frac{\dot{v}}{R} = MgR\sin\gamma\].

Key concepts

Free-body diagram

When approaching problems involving rolling motion, a free-body diagram is a crucial visualization tool. It involves sketching all forces acting on the object, here a solid disk, as it rolls down an inclined plane.
Begin by identifying the key forces:

• **Gravitational Force (\(Mg\))**: This force pulls the disk downward through its center of mass.
• **Normal Force (\(N\))**: Acts perpendicular to the incline, opposing gravity's perpendicular component.
• **Frictional Force (\(f\))**: Acts up the incline at the point of contact, preventing slipping. It's crucial as it enables rolling without slipping.

Diagrams help understand how these forces interact and affect the object's motion. Visualization simplifies the setup for solving mechanical problems by showing the orientation and magnitude of forces involved.

Rotational dynamics

Rotational dynamics revolve around understanding how torque influences rotational motion, much like force affects linear motion.
For a disk rolling about point \(P\), \(\dot{\mathbf{L}}=\Gamma^{\text {ext }}\) signifies that the change in angular momentum equals the external torque on the body.

• **Torque Due to Gravity**: \(MgR\sin\gamma\) is the torque from gravitational force, with gravity trying to rotate the disk.
• **Torque Calculation**: Both normal and frictional forces pass through point \(P\), so they contribute zero torque—only gravity affects rotational motion.

Solving the equation \(\dot{L} = \frac{3}{2}MR^2 \frac{\dot{v}}{R} = MgR\sin\gamma\) helps find linear acceleration, relating rotational inertia and the force applied.

Moment of inertia

Moment of inertia conceptually measures an object's resistance to changes in rotational motion.
For a disk:

• **About Point \(P\)**: The moment of inertia is \(\frac{3}{2} MR^2\), crucial in determining how easily it rotates.
• **About Center of Mass (CM)**: Here, it's \(\frac{1}{2} MR^2\). Different axis leads to altered values.

These values affect the disk's rotation directly. Remember, the larger the inertia, the harder it is to change the object's rotational speed.

Angular acceleration

Angular acceleration (\(\dot{\omega}\)) measures how quickly rotational speed changes over time. It's a vital aspect of any rotational dynamics problem.
In the context of the rolling disk:

• **Non-Slip Condition**: Establishes \(v=R \omega\) and hence \(\dot{v}=R \dot{\omega}\)
• **Stopping and Starting Influence**: Any torque influences this acceleration as it rotates about point \(P\).

By understanding \(\dot{\omega}\), one links linear and rotational motion. It serves as a bridge in understanding how pushing or pulling forces alter spin.