Question

A 3.9 kg ball swings from a string in a vertical circle such that it has constant mechanical energy. Ignore any friction from the air or in the string. What is the difference in the tension between the lowest and highest points on the circle? Give a positive answer in N.

Answer 1

Let the mass of the ball be m = 3.9 kg

Let the speed of the ball at the top of the circle be V1

Let the speed of the ball at the bottom be V2

Let the radius of the circular path in which ball revolves be r

Since air friction is neglected , total energy of the ball is conserved or same at all points of its motion

So it will have same energy at the bottom point and the top

In this case , potential energy is calculated by taking the bottom point that can be reached by the ball as reference point

So , the potential energy at the bottom point is zero.

Kinetic energy of the ball at the bottom is 1/2 m V2²

Potential energy of the ball at the top is mg * ( 2 r ) = 2 mgr

Kinetic energy of the ball at the top is 1/2 m V1²

On using the law of conversation of energy , we get

1/2 mV1² + 2 mgr = 1/2 m V2²

m V1² + 4 mgr = m V2²

( m V1² / r ) + 4 mg = m V2² / r .................. ( 1 )

We will use the above equation later

Now at the top of the circular path , tension will be

T1 = mV1² / r - m g ............. ( 2 )

where the first term on right hand side is centripetal force on the ball when it is at the top

Similarly , tension at the bottom is

T2 = mV2² / r + mg ............... ( 3 )

We can substitute equation ( 1 ) in equation ( 3 )

T2 = mV1² / r + 4 mg + mg

Using equation ( 2 ) , we get ,

T2 = T1 + mg + 4 mg + mg

T2 = T1 + 6 mg

T2 - T1 = 6 mg

= 6 * 3.9 * 9.81

= 229.55 N

So the difference between the tensions at the top and bottom of the circle is 229.55 N