A uniform rod of mass 1.80 kg and length 2.00 m is capable of rotating about an

Question

A uniform rod of mass 1.80 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 = 5.40 kg is attached to one end and a second mass m_2 = 2.70 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 2.40 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 2.40 rad/s? j

Explanation / Answer

here,

mass of rod, mr = 1.80 kg
length of rod, l = 2 m

Distance to mid point of rod, r = L/2 = 2/2 = 1 m

mass of large sphere, m1 = 5.40 kg
mass of small sphere, m2 = 2.70 kg

Part A:
Moment if inertia of system about axis through centre of rod, I
I = inertia of sphere1 + inertia of sphere2 + inertia of rod

I = m1*(r)^2 + m2*(r)^2 + (1/12)*mr*r^2

I = 5.40*(1)^2 + 2.70*(1)^2 + (1/12)*1.80*(1)^2

I = 8.25 Kg.m^2

Part b:
Kinetic Energy, Ke = 0.5 * Net mass * w^2*(l/2)^2 ( w is angular speed)
Ke = 0.5 * (1.80 + 5.40 + 2.70 ) * 2.40^2 * (2/2)^2
Ke = 28.512 J

Part C:
Mass of rod , mr = 0
I = m1*(r)^2 + m1*(r)^2
I = 5.40*(1)^2 + 2.70*(1)^2
I = 8.1 kg.m^2

Part D:
Ke = 0.5 * Net mass * w^2*(l/2)^2
Ke = 0.5 * (5.40 + 2.70 ) * 2.40^2 * (2/2)^2
Ke = 23.328 J

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