What torque would have to be applied to a rod with a length of $6 m$ $6 m$ and a mass of $3 k g$ $3 kg$ to change its horizontal spin by a frequency $4 H z$ $4 Hz$ over $6 s$ $6 s$ ?

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Answer 1 · 2016-12-23T11:09:48

The torque is $= 37.7 N m$ $=37.7Nm$

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 is $I = \frac{1}{12} m l^{2}$ $I=1/12ml^2$

$= \frac{1}{12} \cdot 3 \cdot 6^{2} = 9 k g m^{2}$ $=1/12*3*6^2= 9 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{4}{6} \cdot 2 π$ $(domega)/dt=(4)/6*2pi$

$= \frac{4}{3} π r a d s^{- 2}$ $=4/3pi rads^(-2)$

So the torque is $τ = 9 \cdot \frac{4}{3} π N m = 12 π N m = 37.7 N m$ $tau=9*4/3pi Nm=12piNm=37.7Nm$

What torque would have to be applied to a rod with a length of 6 m6m6 m and a mass of 3 kg3kg3 kg to change its horizontal spin by a frequency 4 Hz4Hz4 Hz over 6 s6s6 s?