A uniform rod of mass $300 g$ and length $50 c m$ rotates at a uniform angular speed of $2 r a d / s$ about an axis perpendicular to the rod through an end. Calculate $(a)$ the angular momentum of the rod about the axis of rotation, $(b)$ the speed of the centre of the rod and $(c)$ its kinetic energy.

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Solution

$m = 0.3 k g$ and length $l = 0.5 m$ with $ω = 2 r a d / s e c$
Moment of inertia about one end is $I = \frac{m l^{2}}{3} = 0.025 K g m^{2}$
(a) Angular momentum is $J = I ω = .025 \times 2 = .05 K g m^{2} r a d / s e c$
(b) The speed of the center of the rod $v = r ω = \frac{l}{2} ω = 0.5 m / s e c$
(c) Its kinetic energy $E = \frac{1}{2} I ω^{2} = 0.05 J o u l e$

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A uniform rod of mass 300g and length 50cm rotates at a uniform angular speed of 2rad/s about an axis perpendicular to the rod through an end. Calculate (a) the angular momentum of the rod about the axis of rotation, (b) the speed of the centre of the rod and (c) its kinetic energy.

m=0.3kg and length l=0.5m with ω=2rad/secMoment of inertia about one end is I=ml23=0.025Kgm2(a) Angular momentum is J=Iω=.025×2=.05Kgm2rad/sec(b) The speed of the center of the rod v=rω=l2ω=0.5m/sec(c) Its kinetic energy E=12Iω2=0.05Joule