What torque would have to be applied to a rod with a length of $3 m$ $3 m$ and a mass of $1 k g$ $1 kg$ to change its horizontal spin by a frequency of $6 H z$ $6 Hz$ over $2 s$ $2 s$ ?

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Answer 1 · 2017-02-10T09:23:26

The torque is $= 14.1 N m$ $=14.1Nm$

The torque is the rate of change of angular momentum

$τ = \frac{d L}{d t} = \frac{d (I ω)}{d t} = I \frac{d ω}{d t}$ $tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt$

The moment of inertia of a rod is $I = \frac{1}{12} \cdot m L^{2}$ $I=1/12*mL^2$

$= \frac{1}{12} \cdot 1 \cdot 3^{2} = \frac{3}{4} k g m^{2}$ $=1/12*1*3^2= 3/4 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{6}{2} \cdot 2 π$ $(domega)/dt=6/2*2pi$

$= (6 π) r a d s^{- 2}$ $=(6pi) rads^(-2)$

So the torque is $τ = \frac{3}{4} \cdot (6 π) N m = \frac{9}{2} π N m = 14.1 N m$ $tau=3/4*(6pi) Nm=9/2piNm=14.1Nm$

What torque would have to be applied to a rod with a length of 3 m3m3 m and a mass of 1 kg1kg1 kg to change its horizontal spin by a frequency of 6 Hz6Hz6 Hz over 2 s2s2 s?