Rotational motion - Rolling without slipping

Problem Statement:

A solid homogeneous disk of mass M and radius R descends an inclined plane while rolling without slipping. The coefficient of friction between the disk and the plane is μ = 0.15. Determine the maximum angle θ for the disc to roll without slipping.

Givens: The moment of inertia of a disk with respect to an axis that passes through its center of mass is ICM = (1/2)MR2

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Solution:

To solve this problem we must bear in mind that the friction force has a different value when the disc rolls than when it slides. We will first determine it for the first situation.

The motion of the disc is the composition of the translation of its center of mass with respect to a reference system at rest and a rotation with respect to an axis that passes through its center of mass. Therefore to describe its motion we will need Newton’s second law (to describe the translation) and the equation of rotational motion:

We start by drawing the external forces that act on the disk. Since it is homogeneous, the center of mass is in the center of the disk and the weight is applied in the center of mass. We have represented the Cartesian axes that we will use to calculate the projections of the different vectors in the figure.

Newton’s second law applied to the center of mass of the disk is:

Then we project it on the axes. If you want to see in more detail how the projections are determined you can consult the problem 6 of Newtonian dynamics.

Let’s write now the equation of the rotational motion applied to the disk. Remember that the torque (or moment) of a force is given by:

Where r is a vector that goes from the point we use as the origin to calculate the torques (in this problem the center of mass of the disk) to the point of application of the force.

The equation of the rotation applied to the disk is:

Torque of the weight P: As the weight is applied in the center of mass of the disk, the point that we have chosen as origin to calculate the torques, r = 0 for the weight and therefore τ P = 0.


Torque of the normal N: The following figure shows the normal force together with its vector r.

Since r and N form an angle of 1800, their cross product (torque) is null.


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Torque of the friction force FR: The torque of the friction force is given by:

Its magnitude is:

Both vectors are represented in the following figure. The direction of the cross product is determined using the right-hand rule. As seen in the figure, the torque vector is directed in the direction –k.

From the previous analysis we deduce that the only force that has a nonzero torque with respect to the center of mass of the disc is the friction force.

Therefore, the equation of the rotational motion for the disk is:

The moment of inertia is a positive scalar; therefore, the angular acceleration vector is parallel to the torque of the friction force. Projecting on the axis z we get:

If the disc rolls without slipping, the rolling condition must be fulfilled:

After substituting it in equation (3) and using the moment of inertia of the disk we obtain the magnitude of the friction force:

On the other hand, from equation (1) we can isolate the acceleration of the center of mass:

And using equation (4) we obtain the magnitude of the friction force as a function of the angle of the plane:

Equation (5) gives the magnitude of the friction force when the disc rolls without sliding. If the angle of the plane is increased, there will come a time when the disk will begin to slide. At that time the friction force will be:

The magnitude of the normal has been obtained from equation (2).

Equalizing the equations (5) and (6) we obtain the angle that the plane must have so that the disk begins to slide:

This will therefore be the maximum angle that the plane must have for the disc to roll without slipping. Note that it does not depend on its mass.

 

 

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