# HOW MUCH WORK IS DONE ON A BODY OF MASS 1 KG WHIRLING ON A CIRCULAR PATH OF RADIUS 5m?

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by ishaRattan

2015-05-20T01:10:54+05:30

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See diagram.

I am not clear on the question really.  Perhaps the following is the expected answer. perhaps not...

A body is whirled in a horizontal circle  ABCD.  The pivot P holds a string to which the stone is tied.  The string is rotated by some means.  A tension develops in the string to balance the weight and  give the stone a centripetal force.

Let the string move on a conical surface of an angle = Ф.  The string makes angle Ф with the vertical.

T Cos Ф = mg    as the body does move vertically upwards or downwards
T Sin Ф = m v² / R

=>  v² = R g Tan Ф
=>  Kinetic energy required :  1/2 m v² = 1/2 m R g Tan Ф
=> Tension required T = mg / CosФ

Initially the body is hanging from the pivot at the end of the string.  So the height to which the body is elevated while whirling = h =  L (1- Cos Ф).

So the increase in PE.  =  m g L (1- CosФ)    , where L is the length of the string.

So the amount work to be done is equal to the KE + PE as above.  Once this Energy is given, the body will continue to move in the circle, as long as the tension in the string is maintained.

Energy required = mg L (1 - Cos Ф) + 1/2 m g R Tan Ф
= m g [ L(1 - CosФ) + 1/2 * R Tan Ф ]

Once, the body is in circular motion, there is no additional work, as  the weight and the tension are both perpendicular to the velocity and displacement of the body.

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If a body is in circular motion (uniform),  horizontally, then there is no work done further to maintain it in circular motion.  This is true in the gravitational field.

If a body makes a circle vertically in the gravitational field,  the the net amount of work done is zero.

In any conservative field of force like  magnetic, electric, gravitation , if a body makes a complete circle, then the net amount of work done is zero.

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