A thin, uniform rod with length of 2m and a linear mass density of 15 kg/m rotat

Question

A thin, uniform rod with length of 2m and a linear mass density of 15 kg/m rotates about an axis perpendicular to its length at one of its ends. What force, applied at an angle of 30 degrees to teh rod a distance of 0.5 m from the axis of rotation as shown in the figure is required to give this rod an angular velocity = of 0.2 rad/sec after it has rotated through one full revolution? Note that the angle of the applied force direction relative to the rod remains constant at 30 degrees thoughout this rotation. The line of force lies in a plane perpendicular to the rotation axis.

Please show step-by-step work and explanation.

Explanation / Answer

We know that
w2 = u2 + 2aS
where w is the final angular velocity , u is the initial angular velocity , a is the angular acceleration ans S is the angular dislacement
u = 0
w = 0.2 , S = 1 rev = 2Pi rad
0.22 = 0 +2a*2Pi
a = 3.183*10-3 rad/s2
we know that
torque = moment of inertia*angular acceleration
torque = force*perpendicular distance (x)
Sin30 = x/0.5
x = 0.25
Force*0.25 = (mL2/3)*3.183*10-3
where m isthe mass of the rod = 15*2 = 30 kg
Force*0.25 = (30*22 /3 )*3.183*10-3
Force = 0.509 N

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