A 3.0-kg rod that is 1.3 m long is free to rotate in a vertical plane about an a

Question

A 3.0-kg rod that is 1.3 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.

Explanation / Answer

a) apply from net torque T T 2- T1 as

torque = torque 2 - torque 1

Tnet = 2*9.81*1.3/2*cos 30 - 1*9.81*1.3 /2*cos 30

=1*9.81*0.7/2*cos(30 degrees)] = 5.522

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b)from the formula of Torque = I alpha

Here Inet = I rod + I mass1 + I mass2

= 1/12*3*1.3^2 + 2*(1.3/2)^2 + 1*(1.3/2)^2

Inet = 1.69

alpha = torque//I = 5.5221.69 = 3.26 rad/s^2

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