Question

A boy standing at the centre of turntable with his arms outstretched is set into rotation with angular speed w rev/min when a boy fold his arms back his moment of inertia reduces to 2/times its initial value find the ratio of his final kinetic energy of rotation to initial kinetic energy

Answer 1

Answer:

Given:

A rotating man is initially with angular Velocity ω. Now he folds his arm and Moment of Inertia reduced to ½ of it's initial value.

To find:

Ratio of New KE to Initial KE

Concept:

When no external torque is applied, the angular momentum of any body remains constant

∴ L = constant

=> I × ω = constant

Calculation:

Let initital Moment of Inertia be I , final Moment of Inertia is I/2.

Let new angular Velocity be ω2

∴ I × ω = (I/2) × ω2

=> ω2 = 2ω ..........(1)

Now initial KE = ½Iω²..........(2)

Final KE = ½ × (I/2) × (2ω)²

=> Final KE = 2 [ ½Iω²] ........(3)

∴ So required ratio :

⇒ 2 [ ½Iω²] : [½Iω²]

⇒ 2 : 1

So, final answer is 2 : 1

Answer 2

QUESTION :-

A boy standing at the centre of turntable with his arms outstretched is set into rotation with angular speed w rev/min when a boy fold his arms back his moment of inertia reduces to 2/times its initial value find the ratio of his final kinetic energy of rotation to initial kinetic energy.

FIND:-

find the ratio of his final kinetic energy of rotation to initial kinetic energy.

ANSWER:-

∴ L = constant

L× ω = constant

HERE,

initital Moment of Inertia is L.

and final Moment of Inertia is L/2.

Let new angular Velocity be ω2

∴ I × ω = (I/2) × ω2

ω2 = 2ω ..........(1)

here,

initial KE = ½Lω²..........(2)

Final KE = ½ × (I/2) × (2ω)²

= Final Kinetic energy = 2 [ ½Lω²] ........(3)

∴ RATIO IS:-

= 2 [ ½Iω²] :1 { [ ½Iω²] }....(FROM EQ. 2& 3).

= 2 : 1

HENCE,

2:1 the ratio of his final kinetic energy of rotation to initial kinetic energy.