First slide

Pure rotational motion and its kinetic energy

Question

A thin disc rotates about an axis passing through its centre and perpendicular to its plane with a constant angular velocity ' ω' I is the moment of inertia of that disc and 'L' is its angular momentum about the given axis. Then rotational kinetic energy of the disc 'E' is
A) E $\large \propto$ L²B) E $\large \propto$ L^-2C) E $\large \propto$ l D)E $\large \propto$ l^-1

Easy

A

Only C are correct
B

B and C are correct
C

B and D are correct
D

A and D are correct

Solution

(KE)_r = 1/2 IW² = 1/2 ( (IW)² / I )
⇒ (KE)_r = L² / 2I
.: E = 1/2 IW² and E = L² / 2I
Also , L = I W
Since W is constant ,
E α i and L α I only (c) is correct.

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