Question

# A 3.00-kg rod that is 2.60 m long is free to rotate in a vertical plane...

A 3.00-kg rod that is 2.60 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 1.00-kg block is attached to one end of the rod, and a 2.00-kg block is attached to the other end. At some instant, the rod makes an angle of 31.0 ? 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.   g = 9.81 m/s2 .

Determine the torque about the pivot 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. When entering the units, use the menu to enter a dot (for multiplication) between factors in the units.

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.

Ans:-

Given data

L= 2.6m, m1= 1kg, m2 = 2kg, ?= 31deg

1]
Only the masses on the ends cause a net torque as each half of the bar is exactly counterbalanced by the opposite end.
T= Fg1*L/2 cos31- Fg2 *L/2 cos31
? = 1.0(9.8)(2.6 / 2)cos31 - 2.0(9.8)(2.6 / 2)cos31
? = 10.9203-21.84
? = -10.92 N•m

Part B

Find the moment of inertia of the system as the sum of the two point masses and the rod moments.

I = m?r² + m?r² + m?L²/12

I = 1.0(2.6 / 2)² + 2.0(2.6 / 2)² + 3.0(2.6)²/12

I= 1.69 + 3.38 + 1.69

I = 6.76 kg•m²

? = I?
? = ?/I

? = -10.92 / 6.76
? = - 1.615

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