What torque would have to be applied to a rod with a length of $7 m$ $7 m$ and a mass of $12 k g$ $12 kg$ to change its horizontal spin by a frequency $14 H z$ $14 Hz$ over $9 s$ $9 s$ ?

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Answer 1 · 2017-01-07T05:57:10

The torque is $= 478.9 N m$ $=478.9Nm$

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

$= \frac{1}{12} \cdot 12 \cdot 7^{2} = 49 k g m^{2}$ $=1/12*12*7^2= 49 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{14}{9} \cdot 2 π$ $(domega)/dt=(14)/9*2pi$

$= (\frac{28 π}{9}) r a d s^{- 2}$ $=((28pi)/9) rads^(-2)$

So the torque is $τ = 49 \cdot \frac{28 π}{9} N m = 478.9 N m$ $tau=49*(28pi)/9 Nm=478.9Nm$

What torque would have to be applied to a rod with a length of 7m7 m and a mass of 12kg12 kg to change its horizontal spin by a frequency 14Hz14 Hz over 9s9 s?