Question

Q11

A body is revolving in a vertical circle with constant mechanical energy. The speed of the body at the highest point is √2rg. The speed of the body at the lowest point is:
O √7gr
О√6gr
O √8gr
O √9gr​

Answer 1

Answer:

The speed of the body at the lowest point is √(6gr)

Explanation:

Let the velocity at the lowest point be "u"

Given,

The speed of the body at the highest point is "v" = $\sqrt{2rg}$

The radius of the vertical circle is "r"

To find,

The velocity "u"

Concept,

Vertical motion :

A body revolves in a vertical circle such that its motion at different points is different, then the motion of the body is stated to be vertical circular motion.

Calculation,

From the conservation of energy,

K.E at the lowest point = K.E at the highest point + P.E at that point

i.e.

$\frac{1}{2}mu^2 = \frac{1}{2} mv^2 + mg(2r)\\\implies u^2 = v^2 + 4gr\\\implies u^2 = 2rg + 4rg\\\implies u = \sqrt{6rg}$

Therefore, the speed of the body at the lowest point is √(6gr)

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