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Question: The string of pendulum of length \(l\) is displaced through \({90^ \circ }\) from the vertical and r...

The string of pendulum of length ll is displaced through 90∘{90^ \circ } from the vertical and release. Then the minimum strength of the string in order to withstand the tension as the pendulum passes through the mean position is:
(a) 5mg
(b) 6mg
(c) mg
(d) 3mg

Explanation

Solution

In this question we have to apply the concept of the body in pendulum motion and we have to utilize the concept of centrifugal force and centripetal force. And the tangential and radial acceleration of the bob of the pendulum.

Complete step by step answer: When we consider the motion of a pendulum. We have to consider the mean position in this question.
In the mean position we have to calculate the tension in the string. Therefore, if we consider the vertical forces acting on the bob at mean position, we know there are three forces acting on it.
The upward centripetal force (mv2r)\left( {\dfrac{{m{v^2}}}{r}} \right), the upward tension of the string (T)(T)and the weight of the body in downward direction (mg)(mg).
Since the body is at vertical equilibrium at mean position, we can form an equation such as:
T−mv2l=mgT - \dfrac{{m{v^2}}}{l} = mg, since r=l.
Therefore, for finding the Tension we have to find velocity first.
We know that bob when at mean position has kinetic energy equal to the potential energy of the bob.
If a bob is attached with a string of length l then the potential energy of the bob is:
U=mgh ⇒U=mgl\begin{array}{l} U = mgh\\\ \Rightarrow U = mgl \end{array}
And the kinetic energy given at that point is:
K=12mv2K = \dfrac{1}{2}m{v^2}
Now we know that these two are equal,
12mv2=mgl ⇒v2=2gl ⇒v=2gl\begin{array}{l} \dfrac{1}{2}m{v^2} = mgl\\\ \Rightarrow {v^2} = 2gl\\\ \Rightarrow v = \sqrt {2gl} \end{array}
Now we have the velocity of bob at the mean position.
We will put the value in the relation we obtained earlier:
T−mv2l=mg ⇒T=mg+mv2l ⇒T=mg+m(2gl)2l ⇒T=mg+m×2gll ⇒T=mg+2mg ⇒T=3mg\begin{array}{l} T - \dfrac{{m{v^2}}}{l} = mg\\\ \Rightarrow T = mg + \dfrac{{m{v^2}}}{l}\\\ \Rightarrow T = mg + \dfrac{{m{{\left( {\sqrt {2gl} } \right)}^2}}}{l}\\\ \Rightarrow T = mg + \dfrac{{m \times 2gl}}{l}\\\ \Rightarrow T = mg + 2mg\\\ \Rightarrow T = 3mg \end{array}
Therefore, the required value of tension is 3mg and the correct option is (d).

Note: In this type of question the imagination of free body diagrams is very important for students. Then it will be easier for them to understand the force acting on a body in a particular axis and summation of which forces are zero.