Rotational motion - Reaction forces and angular acceleration

Problem Statement:

A homogeneous bar of mass M = 15 kg and length L = 6 m is fastened to a vertical wall by an articulation at point A, and supported on a box at point B, at a distance L ‘= 5 m from the wall. The bar is inclined at an angle θ = 300 (see figure).

  1. Calculate the normal force exerted by the box on the bar so that it is in static equilibrium.
  2. Calculate the components of the joint reaction in A.
  3. If the box is removed, determine the angular acceleration of the bar as a function of the angle θ.

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

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Solution:

We are going to solve sections (a) and (b) of the problem by imposing the two conditions of a static equilibrium on the bar:

You can consult more static problems in this link to see in detail how they are solved.

The first condition implies that the bar has no translational motion (a = 0) and the second one that it has no rotational motion (α = 0).

To apply them we will draw the external forces that act on the bar:

Since the bar is homogeneous, its center of mass is located at half its length and its weight (the gravitational force that the Earth exerts on it) is applied at that point.

At point A the bar is attached to the wall by a joint. The reaction in this type of support has two components (Rx and Ry) that are calculated independently.

At point B the support is simple and the reaction (or normal force) is perpendicular to the bar.

In the previous figure we have also represented the three unit vectors that define the positive direction of the Cartesian axes that we will use to project the different vectors.

The first condition of static equilibrium applied to the bar is:

And projecting on the axes we get:


The second condition of static equilibrium must be respected independently of the point chosen to calculate the moment of the forces. In this problem we will use point A. The condition for the bar not to rotate with respect to this point is:

The moment of a force is defined as:

Where r is a vector that goes from the origin of moments (point A) to the point of application of force. this vector is null for the components of the reaction in the joint.


We will now calculate the torque of the weight and of the normal N.

Weight P: The torque of the weight is given by:

Both vectors are represented in the following figure.

The direction of the cross product is given by the right-hand rule. In the figure you can see that the torque vector is directed in the negative direction of the axis z (-k). The magnitude of the vector product is given by:


Normal N: The torque of the normal is given by:

Both vectors are represented in the following figure.

The direction of the cross product is given by the right-hand rule. In the figure you can see that the torque vector is directed in the positive direction of the axis z (k). The magnitude of the vector product is given by:


You can see below the two torques represented on the Cartesian axes:

With this information we are able to project the second condition of static equilibrium on the axis z.

From equation (3), by isolating and replacing with the givens, we get N:

From equation (1) we get Rx:

From equation (2) we get Ry:


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In section (c) of the problem the bar is no longer in static equilibrium; therefore we must use Newton’s second law for rotation to determine its angular acceleration α:

As the axis of rotation of the bar passes through the point A, we will calculate its moment of inertia and the moment of the external forces with respect to this point.

The external forces acting on the bar are represented in the next figure. Note that, since the bar is no longer supported by the box, the normal N does not act on it.

The equation of the rotational motion applied to the bar is then:

As we saw previously, the torque of the components of the reaction in the joint is zero with respect to point A. Therefore the equation is simply:

The moment of inertia is a positive quantity, so the angular acceleration of the bar is parallel to the torque of the weight. We have already seen this torque is directed in the direction –k and has a magnitude:

So the projection on the z axis of the rotational motion equation is:

Using Steiner’s theorem we calculate the moment of inertia of the bar with respect to an axis that passes through point A:

By substituting in the previous equation we get:

And by substituting the givens of the problem statement we get:

In the problem we have used g = 10 m/s2.

Do not forget to include the units in the results.

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