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 $2 H z$ $2 Hz$ over $5 s$ $5 s$ ?

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Answer 1 · 2017-01-29T16:03:54

The torque is $= 22.62 N m$ $=22.62Nm$

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

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

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{2}{5} \cdot 2 π$ $(domega)/dt=(2)/5*2pi$

$= (\frac{4 π}{5}) r a d s^{- 2}$ $=((4pi)/5) rads^(-2)$

So the torque is $τ = 9 \cdot \frac{4 π}{5} N m = \frac{36}{5} π N m = 22.62 N m$ $tau=9*(4pi)/5 Nm=36/5piNm=22.62Nm$

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 2 Hz2Hz2 Hz over 5 s5s5 s?