A uniform rod is set up so that it can rotate about a perpendicular axis at one

Question

A uniform rod is set up so that it can rotate about a perpendicular axis at one of its ends. The length and mass of the rod are 0.845 m and 1.31 kg. What constant-magnitude force acting at the other end of the rod perpendicularly both to the rod and to the axis will accelerate the rod from rest to an angular speed of 6.93 rad/s in 8.23 s?
A uniform rod is set up so that it can rotate about a perpendicular axis at one of its ends. The length and mass of the rod are 0.845 m and 1.31 kg. What constant-magnitude force acting at the other end of the rod perpendicularly both to the rod and to the axis will accelerate the rod from rest to an angular speed of 6.93 rad/s in 8.23 s?
A uniform rod is set up so that it can rotate about a perpendicular axis at one of its ends. The length and mass of the rod are 0.845 m and 1.31 kg. What constant-magnitude force acting at the other end of the rod perpendicularly both to the rod and to the axis will accelerate the rod from rest to an angular speed of 6.93 rad/s in 8.23 s?
A uniform rod is set up so that it can rotate about a perpendicular axis at one of its ends. The length and mass of the rod are 0.845 m and 1.31 kg. What constant-magnitude force acting at the other end of the rod perpendicularly both to the rod and to the axis will accelerate the rod from rest to an angular speed of 6.93 rad/s in 8.23 s?

Explanation / Answer

the length of the rod l = 0.845 m

mass of the rod is m = 1.31 kg

final angular speed is wf = 6.93 rad/s

initial angular speed is wo = 0

the time is t = 8.23 s

let alpha be the angular acceleration

we know that

wf = wo + alpha x t

or alpha = (wf - wo/t)

the tangential acceleration is

a_t = r x alpha

where r = (l/2)

the force acting at the other end is

F = m x a_t

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