A 75.0 gram uniform density rod of length o.800 m is rotating counterclockwise a

Question

A 75.0 gram uniform density rod of length o.800 m is rotating counterclockwise about an axis that is 0.320 m from the center of the rod, with an angular speed of 32.0 rad/s. We apply a 0.600 N force at the end furthest away from the axis, creating a clockwise torque that slows the rod to a stop. a.) What is the moment of inertia of the rod about this axis? b.) Calculate the clockwise torque being applied to the rod. c.) Calculate the angular acceleration of the rod due to the clockwise torque. d.) Calculate the time it takes to slow the rod to a stop.

Explanation / Answer

let m = 75 g = 0.075 kg

L = 0.8 m

d = 0.32 m

a) Moment of Inertia about the axis of rotation,

I = m*L^2/12 + m*d^2 (using parallel axis theorem)

= 0.075*0.8^2/12 + 0.075*0.32^2

= 1.168*10^-2 kg.m^2

b) clockwise torque, T = F*(L/2 + d)*sin(90)

= 0.6*(0.8/2 + 0.32)*1

= 0.432 N.m

c) angular acceleration of the rod, alfa = -T/I

= -0.432/(1.168*10^-2)

= -37 rad/s^2

d) let t is the time taken to stop.

use, wf = wi + alfa*t

==> t = (wf - wi)/alfa

= ( 0 - 32)/(-37)

= 0.865 s

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