A uniform rod of mass 1.60 kg and length 2.00 m is capable of rotating about an

Question

A uniform rod of mass 1.60 kg and length 2.00 m is capable of rotating about an axis passing through its center and perpendicular to its length. A mass m1 = 5.00 kg is attached to one end and a second mass m2 = 2.00 kg is attached to the other end of the rod. Treat the two masses as point particles.

(a) What is the moment of inertia of the system?

(b) If the rod rotates with an angular speed of 2.10 rad/s, how much kinetic energy does the system have?

(c) Now consider the rod to be of negligible mass. What is the moment of inertia of the rod and masses combined?

(d) If the rod is of negligible mass, what is the kinetic energy when the angular speed is 2.10 rad/s?

Explanation / Answer

moment of inertia

I = (1/12)*m*l^2 + m1*(l/2)^2 + m2*(l/2)^2

I = (1/12)*1.6*2^2 + (5*1^2) + (2*1^2)

I = 7.53 kg m^2

==================

(b)

kinetic energy KE = (1/2)*I*w^2

KE = (1/2)*7.53*2.1^2 = 16.6 J

=================

(c)

moment of inertia I = m1*(l/2)^2 + m2*(l/2)^2

I = (5*1^2) + (2*1^2) = 7 kg m^2

===================

kinetic energy KE = (1/2)*I*w^2

KE = (1/2)*7*2.1^2 = 15.435 J

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