Rotational motion - Angular acceleration of a pulley

Problem Statement:

A homogeneous pulley with two grooves consists of two wheels which turn together as one around the same axis. The moment of inertia of the two wheels together is ICM = 40 kg m2. The radii are: R1 = 1.2 m and R2 = 0.4 m. The masses that hang on both sides of the pulley are m1 = 36 kg and m2 = 12 kg (see figure). We will assume that the masses of the ropes are negligible.

Determine the angular acceleration of the pulley and the tensions of the ropes.

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Solution:

The system represented in the figure is constituted by three bodies: the pulley and the two masses. To describe the motion of each of them, we must use the appropriate equation according to the type of motion they describe.

The masses describe a motion of translation, so their linear accelerations are given by Newton’s second law:

The pulley describes a rotational motion, so its angular acceleration will be given by Newton’s second law for rotation:

First we draw the forces that act on the system:

Note that the tension in each rope is different. The weight and the normal to the axis that holds the pulley in place are two other forces that act on it. But these two forces cancel out (they have the same magnitude and are in opposite directions) so they do not affect the motion of the pulley.

In the figure, we have also represented the positive direction of the y axes that we will use to calculate the force projections for each of the masses. We chose to orient the y axis in the same direction as the acceleration vector for each of the masses. The projection of acceleration will be positive in this way.

Note that the masses have different accelerations. This is because each rope is at a different distance from the center of the pulley. We will come back to this later.

We will now write the equation of motion of each body.

Mass m1: Newton’s second law applied to m1 is:

And projecting on the axis y we get:


Mass m2: Newton’s second law applied to m2 is:

And projecting on the axis y we get:


Pulley: Newton’s second law for rotation applied to the pulley is:

The torque (or moment) of a force is given by:

Where r is a vector that goes from the point we choose as the origin of the moments to the point of application of the force.

The direction of the cross product is determined using the right-hand rule, and the magnitude is:

Where θ is the angle that form both vectors.

We will now determine the torques of the tensions to develop the equation of the moments.

In the following figure are represented the pulley and the tensions that act on it:

The torque of T1 is given by:

Its direction, as seen on the left of the figure, is given by the right-hand rule. The torque vector is perpendicular to the plane of the screen and outward.

Its magnitude is given by:

The torque of T2 is given by:

Its direction, as seen on the right of the figure, is given by the right-hand rule. The torque vector is perpendicular to the plane of the screen and inward.

Its magnitude is given by:

In the following figure the two vectors are represented on the Cartesian axes.

The angular acceleration vector α of the pulley is directed in the positive direction of the z axis because the pulley rotates counterclockwise.

Next, we project Newton’s second law for rotation on the axis z:

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Equations (1), (2) and (3) will allow us to solve the problem.

In addition, we can relate the linear acceleration of each mass with the angular acceleration of the pulley:

Since the ropes do not slide on the pulley and therefore describe a circular motion of acceleration α.

Substituting the previous expressions in equations (1) and (2) we finally have the system of equations:

After solving the system of equations and replacing the givens of the statement we get:

In the problem we have used g = 10 m / s 2.

 

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