Determine the velocity of the mass of a simple pendulum of length l at the lowest point of its trajectory, as well as the tension of the string at this same point. The initial angle of the pendulum string with the vertical is α0.
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Solution:
We will start by drawing the forces acting on the mass of the pendulum at any point of its trajectory. The forces are the tension of the string and the weight (if it is subject to the gravity of the Earth).
The displacement vector (in blue) is also represented in the figure above, it is always tangent to the trajectory.
The initial and final states that we will use to apply the energy conservation principle are represented in the figure below. We have also included the zero height position.
The tension of the string is always perpendicular to the displacement, so it does not produce work:
The other force acting on the mass, its weight, is conservative. So the mechanical energy of the mass will be conserved during its motion.
At point A the mass of the pendulum has only potential energy because it is at rest. At point B it has only kinetic energy because its height is hB = 0.
We can then write the energy conservation principle:
We must write the height hA according to the givens of the problem; that is, as a function of the length l of the pendulum and the initial angle α0.
We will use the figure below to write hA in terms of the givens of the problem.
From the previous figure we deduce:
After substituting in the energy conservation equation we get:
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We use Newton’s second law to calculate the tension of the rope at point B. The forces that act on the mass are the tension and the weight. They are represented in the figure below. We have also represented the axes that we will use to project the vectors.
Newton’s second law for the mass at point B is:
And by making the projection on the n-axis (the forces have a null projection on the t-axis at this point) we obtain:
The normal acceleration appears in the second member as it is by definition the projection of the acceleration vector on the axis perpendicular to the trajectory.
To finish, we substitute the value of the velocity at point B (calculated previously) in equation (1) and we isolate T:
It is important that you notice that the tension at point B and the weight cannot have the same magnitude. It must indeed have a higher magnitude than the weight. If both forces canceled each other the mass could not have a normal acceleration because the second member of equation (1) would be zero. The mass must have a normal acceleration so that it describes a curvilinear trajectory.
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