A 3.0-kg rod that is 0.9 m long is free to rotate in a vertical plane about an a

Question

A 3.0-kg rod that is 0.9 m long is free to rotate in a vertical plane about an axle that runs through the rod's center, is perpendicular to the rod's length, and runs parallel to the floor. A 1.0-kg block is attached to one end of the rod, and a 2.0-kg block is attached to the other end. At some instant, the rod makes an angle of 30? with the horizontal so that the blocks are in the positions shown in (Figure 1) . Ignore friction and assume the blocks are small enough that any length they add to the rod can be ignored.

A)Determine the torque caused by the forces exerted on the system at this instant.? Express your answer with the appropriate units. Enter positive value if the torque is counterclockwise and negative value if the torque is clockwise.

B)Determine the angular acceleration of the system at this instant. Enter positive value if the angular acceleration is counterclockwise and negative value if the angular acceleration is clockwise.

Explanation / Answer

Mass of rod m= 3 kg

Length of rod L = 0.9 m

m1 = 1 kg

m2 = 2 kg

Moment of inertia of system about the given axis

I = (mL2/12 ) + (m1L2 /4) + (m2 L2/4) = 0.81 kg.m2

Angle made bby rod from horizontal = 30°

(a). Torque = 2×9.8×cos(30°)×0.45 - 9.8×cos(30°)×0.45 = 3.82 Nm

I assumed that 1 kg is in right , otherwise you can put negative sign.

(b). Angular acceleration = torque/I = 3.82/0.81 = 4.72 rad/s2

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