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' isA) E L2 B) E L-2 C) E l D)E l-1
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a
Only C are correct
b
B and C are correct
c
B and D are correct
d
A and D are correct
answer is A.
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Detailed Solution
(KE)r = 1/2 IW2 = 1/2 ( (IW)2 / I )⇒ (KE)r = L2 / 2I.: E = 1/2 IW2 and E = L2 / 2IAlso , L = I WSince W is constant ,E α i and L α I only (c) is correct.
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' isA) E L2 B) E L-2 C) E l D)E l-1