First slide

Law of conservation of angular momentum

Question

Two discs of same moment of inertia rotating about their regular axis passing through centre and
perpendicular to the plane of disc with angular velocities $ω_{1}$ and $ω_{2}$ . They are brought into contact
face to face coinciding the axis of rotation. The expression for loss of energy during this process is
[NEET 2017]

Moderate

A

$\frac{1}{2} I {(ω_{1} + ω_{2})}^{2}$
B

$\frac{1}{4} I {(ω_{1} - ω_{2})}^{2}$
C

$I {(ω_{1} - ω_{2})}^{2}$
D

$\frac{I}{8} {(ω_{1} - ω_{2})}^{2}$

Solution

When no external torque acts on system, then angular
momentum of system remains constant.
Angular momentum before contact
$= I ω_{1} + I_{2} ω_{2} = I ω_{1} + I ω_{2} = I (ω_{1} + ω_{2})$
Angular momentum after the discs brought into contact
$= I_{net} ω = (I_{1} + I_{2}) ω = 2 I ω$
From law of conservation of angular momentum,
$I (ω_{1} + ω_{2}) = 2 I ω \Rightarrow ω = \frac{ω_{1} + ω_{2}}{2}$
Now, to calculate loss of energy, we subfract initial and final energies of system.
Loss of energy $= \frac{1}{2} I ω_{1}^{2} + \frac{1}{2} I ω_{2}^{2} - \frac{1}{2} (2 I) ω^{2} = \frac{1}{4} I {(ω_{1} - ω_{2})}^{2}$

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