AP EAPCET 07-Jul-2022 Shift 1 Paper

Let,
I1= Inertia of first (rotating) wheel
ω1= Angular speed of first wheel
I2=3I1= Inertia of second wheel
ω2=0= Angular speed of second wheel let
ω= Angular speed of coupled wheels.
According to conservation of angular momentum, total initial angular momentum = total final angular momentum
⇒I1ω1+I2×0=(I1+I2)ω
⇒ω=

I1ω1

I1+I2

=

I1ω1

I1+3I1

⇒ω=

ω1

4

Now, initial rotational energy of system is
K1=

1

2

I1ω12+

1

2

I2(0)2
⇒K1=

1

2

I1ω12
Final rotational kinetic energy of system is
K2=

1

2

(I1+I2)ω2
=

1

2

(I1+3I1)⋅

ω12

16

or K2=

1

4

K1
=

1

2

×4I1×

ω12

16

=

1

4

(

1

2

I1ω12)
Hence, fraction of lost energy
=

∆K

K1

=

K1−K2

K1

=

3 4 K1

K1

=

3

4

=0.75