Question

Explain how the energy is conserved in the case of a body executing circular motion in a vertical plane. If the speed of the body at the highest point is 3 ms-1, find its speed at the lowest point, when the radius of the circular path is 1 m. (Take g = 10 ms-2)

Answer 1

Given:

A body executing circular motion in a vertical plane. If the speed of the body at the highest point is 3 ms-1.

To find:

• Explaination of how energy is conserved during vertical circular motion.

• Speed at lowest point.

Solution:

1) Explaination: A body undergoing motion in a vertical circle obeys the law of conservation of mechanical energy such that the sum of kinetic energy and potential energy at every point will be equal and constant.

2) Calculation:

Let Velocity at lowest point be v ;

Applying Conservation of Mechanical Energy:

$\dfrac{1}{2} m {u}^{2} + mg(2l) = \dfrac{1}{2} m {v}^{2} + 0$

$= > \dfrac{1}{2} m {(3)}^{2} + mg(2 \times 1) = \dfrac{1}{2} m {v}^{2} + 0$

$= > \dfrac{1}{2} m (9) + mg(2) = \dfrac{1}{2} m {v}^{2}$

$= > \dfrac{1}{2} \cancel{m} (9) + \cancel{m}g(2) = \dfrac{1}{2} \cancel{m }{v}^{2}$

$= > \dfrac{9}{2} + 2g = \dfrac{ {v}^{2} }{2}$

$= > 9 + 4g = {v}^{2}$

$= > 9 + 40 = {v}^{2}$

$= > 49 = {v}^{2}$

$= > {v}^{2} = 49$

$= > v = 7 \: m {s}^{ - 1}$

So, final answer is:

$\boxed{ \sf{ \red{v = 7 \: m {s}^{ - 1} }}}$