A uniform rod of mass 2.10 kg and length 2.00 m is capable of rotating about an

Question

A uniform rod of mass 2.10 kg and length 2.00 m is capable of rotating about an axis passing through its center and perpendicular to its length. A mass m_1 = 4.50 kg is attached to one end and a second mass m_2 = 2.80 kg is attached to the other end of the rod. Treat the two masses as point particles. What is the moment of inertia of the system? Kg. m^2 If the rod rotates with an angular speed of 2.30 rad/s, how much kinetic energy does the system have? J Now consider the rod to be of negligible mass. What is the moment of inertia of the rod and masses combined? Kg. m^2 If the rod is of negligible mass, what is the kinetic energy when the angular speed is 2.30 rad/s? J

Explanation / Answer

a)
I of rod = m*L^2/12

Total I = m*L^2/12 + m1*(L/2)^2 + m2*(L/2)^2
= 2.1*2^2/12 + 4.5*(2/2)^2 + 2.80*(2/2)^2
= 0.7 + 4.5 + 2.8
= 8 Kgm^2

b)
KE = 0.5*I*w^2
= 0.5*8*(2.3)^2
=21.16 J

c)
Total I = m1*(L/2)^2 + m2*(L/2)^2
= 4.5*(2/2)^2 + 2.80*(2/2)^2
= 4.5 + 2.8
= 7.3 Kgm^2

d)
KE = 0.5*I*w^2
= 0.5*7.3*(2.3)^2
=19.31 J

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