Rotational motion - Angular acceleration of a rod

Problem Statement:

A homogeneous bar of length L = 2 m is nailed to the wall at its midpoint, so that it can freely rotate around that point. The bar has a mass M = 1 kg, and it is subjected to the following forces: F1 = 4 N, F2 = 4 N, F3 = 12 N (see figure). The distance d = 0.75 m.

Determine the angular acceleration α of the bar as a function of the angle θ assuming that the forces are always vertical.

Givens: the moment of inertia of the bar with respect to an axis that passes through its center of mass is: ICM = (1/12)ML2

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Solution:

To determine the angular acceleration of the bar we must use Newton’s second law for rotation :

Where Στext is the sum of the torques of external forces, ICM is the moment of inertia of the bar with respect to an axis which passes through the center of mass and α the angular acceleration of the bar.

We will calculate the torque of the forces with respect to the center of mass of the bar, since the axis of rotation passes through that point.

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

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

The torque vector magnitude (or norm), like for any other cross product, is:

Where θ is the angle formed by both vectors.

The direction of the cross product is determined with the help of the right-hand rule, as we will see next.

The external forces acting on the bar are represented in the figure included in the problem statement. The weight and the reaction of the support also act on the bar. But since both are applied in the center of mass, the vector r is null and therefore their torque too. Therefore they do not influence the rotational motion of the bar.

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

We will now calculate the torque of each of the forces.

Force F1: In the following figure the force is represented along with its vector r:

The direction of the torque is given by the right-hand rule. As indicated in the figure, the torque vector is perpendicular to the plane of the screen and outward. The torque magnitude is:


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Force F2: In the following figure the force is represented along with its vector r:

The direction of the torque is given by the right-hand rule. As indicated in the figure, the torque vector is perpendicular to the plane of the screen and outward. The torque magnitude is:


Force F3: In the following figure the force is represented along with its vector r:

The direction of the torque is given by the right-hand rule. As indicated in the figure, the torque vector is perpendicular to the plane of the screen and inward. The torque magnitude is:

Finally we represent the three vectors on the Cartesian axes.

The angular acceleration is also shown in the figure. Newton’s second law for rotation defines that this vector must always be parallel to the sum of the torques, since the moment of inertia is a positive scalar. In the figure we have represented it with a negative direction because, as you will see below, in this problem the sign of the angular acceleration is negative.

We can now project Newton’s second law for rotation on the axis z:

And after replacing with the givens of the problem we get:

Do not forget to include the units in the results.

 

 

 

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