What torque would have to be applied to a rod with a length of $8 m$ $8 m$ and a mass of $8 k g$ $8 kg$ to change its horizontal spin by a frequency $7 H z$ $7 Hz$ over $3 s$ $3 s$ ?

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Answer 1 · 2017-01-12T15:20:17

The torque is $625.5 N m$ $625.5Nm$

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 $I = \frac{1}{12} \cdot m L^{2}$ $I=1/12*mL^2$

$= \frac{1}{12} \cdot 8 \cdot 8^{2} = \frac{512}{12} k g m^{2}$ $=1/12*8*8^2= 512/12 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{7}{3} \cdot 2 π$ $(domega)/dt=(7)/3*2pi$

$= (\frac{14 π}{3}) r a d s^{- 2}$ $=((14pi)/3) rads^(-2)$

So the torque is $τ = \frac{512}{12} \cdot \frac{14 π}{3} N m = 625.5 N m$ $tau=512/12*(14pi)/3 Nm=625.5Nm$

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