What torque would have to be applied to a rod with a length of #5 m# and a mass of #2 kg# to change its horizontal spin by a frequency of #7 Hz# over #3 s#?

1 Answer
Jan 22, 2018

Answer:

#244.4 N.m#

Explanation:

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Torque is defined as the rate of change in angular momentum,

i.e #tau = (dL)/dt = (d(Iomega))/dt = I (domega)/dt#

Now,moment of inertia(#I#) of a rod w.r.t its one end is #(Ml^2)/3#,(as here its length is the radius of the circle in which it moves)

so,its #I =16.67 Kg.m^2# (given #l#=5 m and #M#=#2Kg#)

Now,its rate of change in angular velocity is #(domega)/dt = 7/3*2pi# (as, #omega = 2pi*nu# where #nu# is frequency)

So,torque acting is #16.67*(7/3)*2pi# 0r, #244.4 N.m#