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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

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

Step by step solution

01

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 \(P\).
02

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 \[\frac{3}{2}MR^2 \frac{\dot{v}}{R} = MgR\sin\gamma\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

The masses of the earth and sun are \(M_{\mathrm{e}} \approx 6.0 \times 10^{24}\) and \(M_{\mathrm{s}} \approx 2.0 \times 10^{30}\) (both in \(\mathrm{kg}\) ) and their center-to-center distance is \(1.5 \times 10^{8} \mathrm{km}\). Find the position of their \(\mathrm{CM}\) and comment. (The radius of the sun is \(\left.R_{\mathrm{s}} \approx 7.0 \times 10^{5} \mathrm{km} .\right).\)

[Computer] A grenade is thrown with initial velocity \(\mathbf{v}_{\mathrm{o}}\) from the origin at the top of a high cliff, subject to negligible air resistance. (a) Using a suitable plotting program, plot the orbit, with the following parameters: \(\mathbf{v}_{\mathrm{o}}=(4,4), g=1,\) and \(0 \leq t \leq 4\) (and with \(x\) measured horizontally and \(y\) vertically up). Add to your plot suitable marks (dots or crosses, for example) to show the positions of the grenade at \(t=1,2,3,4 .\) (b) At \(t=4,\) when the grenade's velocity is \(\mathbf{v},\) it explodes into two equal pieces, one of which moves off with velocity \(\mathbf{v}+\Delta \mathbf{v} .\) What is the velocity of the other piece? (c) Assuming that \(\Delta \mathbf{v}=(1,3),\) add to your original plot the paths of the two pieces for \(4 \leq t \leq 9 .\) Insert marks to show their positions at \(t=5,6,7,8,9\). Find some way to show clearly that the CM of the two pieces continues to follow the original parabolic path.

A particle of mass \(m\) is moving on a frictionless horizontal table and is attached to a massless string, whose other end passes through a hole in the table, where I am holding it. Initially the particle is moving in a circle of radius \(r_{\mathrm{o}}\) with angular velocity \(\omega_{\mathrm{o}},\) but I now pull the string down through the hole until a length \(r\) remains between the hole and the particle. What is the particle's angular velocity now?

Many applications of conservation of momentum involve conservation of energy as well, and we haven't yet begun our discussion of energy. Nevertheless, you know enough about energy from your introductory physics course to handle some problems of this type. Here is one elegant example: An elastic collision between two bodies is defined as a collision in which the total kinetic energy of the two bodies after the collision is the same as that before. (A familiar example is the collision between two billiard balls, which generally lose extremely little of their total kinetic energy.) Consider an elastic collision between two equal mass bodies, one of which is initially at rest. Let their velocities be \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}=0\) before the collision, and \(\mathbf{v}_{1}^{\prime}\) and \(\mathbf{v}_{2}^{\prime}\) after. Write down the vector equation representing conservation of momentum and the scalar equation which expresses that the collision is elastic. Use these to prove that the angle between \(\mathbf{v}_{1}^{\prime}\) and \(\mathbf{v}_{2}^{\prime}\) is \(90^{\circ} .\) This result was important in the history of atomic and nuclear physics: That two bodies emerged from a collision traveling on perpendicular paths was strongly suggestive that they had equal mass and had undergone an elastic collision.

Use spherical polar coordinates \(r, \theta, \phi\) to find the CM of a uniform solid hemisphere of radius R, whose flat face lies in the \(x y\) plane with its center at the origin. Before you do this, you will need to convince yourself that the element of volume in spherical polars is \(d V=r^{2} d r \sin \theta d \theta d \phi\). (Spherical polar coordinates are defined in Section 4.8. If you are not already familiar with these coordinates, you should probably not try this problem yet.)

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