Question

The bob of a pendulum has mass $m = 1\;{\rm{kg}}$ and charge $q = 40\;{\rm{\mu C}}$. Length of the pendulum is $l = 0.9\;{\rm{m}}$. The point of suspension also has the same charge $40\;{\rm{\mu C}}$. What the minimum speed u should be imparted to the bob so that it can complete a vertical circle?

A. $6\;{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}$
B. $2\;{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}$
C. $8\;{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}$
D. $4\;{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}$

Answer 1

The above problem can be resolved using the concept and application of the fundamentals of the simple pendulum. In this problem, we can use the mathematical formula for the minimum speed that is required to achieve by the bob of the pendulum. Moreover, by substituting the values of the gravitational acceleration and the length of the pendulum, the desired value of the minimum speed can be calculated.

Complete step by step answer:
The mass of the bob of the pendulum is, $m = 1\;{\rm{kg}}$.
The magnitude of charge is,

q = 40\;{\rm{\mu C}}\\\ {\rm{q}} = 40\;{\rm{\mu C}} \times \dfrac{{{{10}^{ - 6}}\;{\rm{C}}}}{{1\;{\rm{\mu C}}}}\\\ q = 40 \times {10^{ - 6}}\;{\rm{C}} \end{array}$$. The length of the pendulum is, $$l = 0.9\;{\rm{m}}$$. We know the expression for minimum speed is given as, $$u = \sqrt {4gl} $$ Here, g is the magnitude of gravitational acceleration and its value is $$9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$$. Solve by substituting the values in the above expression as, $$\begin{array}{l} u = \sqrt {4 \times 9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} \times 0.9\;{\rm{m}}} \\\ u \approx 6\;{\rm{m/s}} \end{array}$$ Therefore, the minimum speed is of $$6\;{\rm{m/s}}$$ and option (A) is correct. **Note:** To resolve the given problem, it is important to remember that the relation for the minimum speed of the bob of the pendulum is correct. In addition to this, the correct knowledge of the variables is required to be taken into consideration. Moreover, the effect of gravity is also needed to be considered while doing such analysis.