What torque would have to be applied to a rod with a length of $7 m$ $7 m$ and a mass of $9 k g$ $9 kg$ to change its horizontal spin by a frequency $15 H z$ $15 Hz$ over $9 s$ $9 s$ ?

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Answer 1 · 2017-02-15T11:15:43

The torque is $= 384.8 N m$ $=384.8Nm$

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

$= \frac{1}{12} \cdot 9 \cdot 7^{2} = \frac{147}{4} k g m^{2}$ $=1/12*9*7^2= 147/4 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{15}{9} \cdot 2 π$ $(domega)/dt=(15)/9*2pi$

$= (\frac{10 π}{3}) r a d s^{- 2}$ $=((10pi)/3) rads^(-2)$

So the torque is $τ = \frac{147}{4} \cdot \frac{10 π}{3} N m = \frac{1470}{12} π N m = 384.8 N m$ $tau=147/4*(10pi)/3 Nm=1470/12piNm=384.8Nm$

What torque would have to be applied to a rod with a length of 7 m7m7 m and a mass of 9 kg9kg9 kg to change its horizontal spin by a frequency 15 Hz15Hz15 Hz over 9 s9s9 s?