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 $9 H z$ $9 Hz$ over $16 s$ $16 s$ ?

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Answer 1 · 2017-01-02T07:05:43

The torque is $= 129.9 N m$ $=129.9Nm$

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

$\frac{d ω}{d t} = \frac{9}{16} \cdot 2 π$ $(domega)/dt=(9)/16*2pi$

$= (\frac{9 π}{8}) r a d s^{- 2}$ $=((9pi)/8) rads^(-2)$

So the torque is $τ = \frac{147}{4} \cdot \frac{9 π}{8} N m = \frac{1323}{32} π N m = 129.9 N m$ $tau=147/4*(9pi)/8 Nm=1323/32piNm=129.9Nm$

What torque would have to be applied to a rod with a length of 7m7 m and a mass of 9kg9 kg to change its horizontal spin by a frequency 9Hz9 Hz over 16s16 s?