A light rod 1.00 m in length rotates about an axis perpendicular to its length & passing through its center?
I am unsure where to begin with this problem! A light rod, measuring 1.00 m in length, rotates about an axis that is perpendicular to its length and passes through its center, as illustrated in the figure below. Two particles with masses of 4.00 kg and 3.00 kg are connected to the ends of the rod.
(a) Neglecting the mass of the rod, what is the kinetic energy of the system when its angular speed is 2.50 rad/s?
(b) Now, repeat the calculation, assuming the mass of the rod is 2.10 kg.
1 Answers
The kinetic energy of a mass M rotating at distance R from an axis with angular velocity w , is
K = (1/2)Iw^2
where I is the mass's moment of inertia, which is just MR^2 if its concentrated at distance R.
So for part (a)
K = (1/2)M1R^2w^2 + (1/2)M2R^2w^2
=(1/2)[M1 + M2]R^2w^2
= (1/2)4 + 3^2(2.5)^2
= 5.47 J
In part (b) you have to add the moment of inertia of the rod which is I = (1/12)ML^2 = (1/12)(2.1)(1)^2 = .175
K = (1/2)M1R^2w^2 + (1/2)M2R^2w^2 + (1/2)Iw^2
= 5.47 + (1/2)(.175)(2.5)^2
= 6 J
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