Question

A bob of mass 100g tied at the end of a string of length 50 cm is revolved in a vertical circle with constant speed of 1m/s. When the tension in the string is 0.7N, the angle made by the vertical is $\left( {g = 10m{s^{ - 2}}} \right)$.
A) ${0^\circ }$
B) ${90^\circ }$
C) ${180^\circ }$
D) ${60^\circ }$

Answer 1

Hint: Here we have to balance the equation. In this three forces are acting on the system one is the downward force due to gravity another one is the upward force due to tension in the string and the third one is the centripetal force acting towards the center.

Complete step by step solution:
Here the net force on the bob is equal to the force acting downwards due to gravity and the tension which is acting opposite to the downward force:
${F_{Net}} = T - mg\cos \theta $;
Here the net force is also equal to the centripetal force which is acting in the center. So:
$\dfrac{{M{v^2}}}{r} = T - mg\cos \theta $;
Put the given value in the above equation:
$\dfrac{{0.1kg \times {1^2}m/s}}{{0.5m}} = 0.7N - 0.1 \times 10\cos \theta $;
$ \Rightarrow \dfrac{{0.1}}{{0.5}} = 0.7 - 0.1 \times 10\cos \theta $;
Do the necessary calculation:
$ \Rightarrow \dfrac{1}{5} - 0.7 = - 0.1 \times 10\cos \theta $
$ \Rightarrow - 0.5 = - 0.1 \times 10\cos \theta $;
Write above equation in terms of$\theta $:
$ \Rightarrow 5 = 10\cos \theta $;
$ \Rightarrow \dfrac{5}{{10}} = \cos \theta $;
Solve,
$ \Rightarrow \dfrac{1}{2} = \cos \theta $;
$ \Rightarrow {\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \theta $;
The angle is:
$ \Rightarrow \theta = 60^\circ $;

Option (D) is correct.

When the tension in the string is 0.7N, the angle made by the vertical is $60^\circ $.

Note: Here we have to equate the net force with the centripetal force. The net force is acting vertically as a downward force and a force in the form of tension is acting upwards. We have to resolve the vectors for the downward force and since the bob is rotated in a vertical circle we will only consider the vertical resolved vector i.e. $\cos\theta $.