Please help me solve it step by step.

Question

A uniform rod rotates in a horizontal plane about a vertical axis through a point which is 2

m away from its center. The rod is 8 m long, weighs 12 N, and rotates at 240 rev/minute.

Calculate

(a) Its rotational inertia about the axis of rotation.

(b) The magnitude of its angular momentum about the axis.

Answer 1

(a) The moment of inertia of a uniform rod about its center of mass is I_0=(1/12)*M*L^2. The length of the rod is given, but rather than the mass, the weight is given. The mass can be calculated by:

(1) F_g = mg => m-(F_g)/(g) = (12N)/(9.81 m/s^2) = 1.223 kg.

From here, the center of mass moment of inertia can be found:

(2) I_0=(1/12)*M*L^2 = (1/12)*(1.223)*(8)^2 = 6.534 kg*m^2

The moment of inertia of an object about any axis can be calculated from its center of mass moment of inertia by the parallel axis theorem: I = I_0 + m*(Δx)^2

Since the actual axis of rotation is 2m away from the center of mass, Δx = 2m, so

(3) I = I_0 + m*(Δx)^2 = 6.534 + 1.223*(2)^2 = 11.42 kg*m^2

The rotational intertia of the rod about its axis of rotation is 11.42 kg*m^2.

(b) The angular momentum (L) of an object can be found from L=I*ω, where ω is the angular velocity in rad/sec. The given rotational speed is in units of revolutions per minute, which must be converted

(4) ω = 240 (rev/min) * (1 min/ 60 sec) * (2π rad/ 1 rev) = 240*2*π/60 = 8π = 25.13 rad/sec

Now the angular momentum can be found from the definition.

(5) L = I*ω = 11.42*25.13 = 286.9 kg*m^2/s

The magnitude of the angular momentum is 286.9 kg*m^2/s