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 $3 s$ ?

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Answer 1 · 2018-01-22T07:15:44

$244.4 N.m$

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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, $244.4 N.m$

What torque would have to be applied to a rod with a length of 5 m5 m and a mass of 2 kg2 kg to change its horizontal spin by a frequency of 7 Hz7 Hz over 3 s3 s?