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 $8 H z$ $8 Hz$ over $7 s$ $7 s$ ?

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Answer 1 · 2017-02-03T11:08:40

The torque is $= 50.3 N m$ $=50.3Nm$

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

$\frac{d ω}{d t} = \frac{8}{7} \cdot 2 π$ $(domega)/dt=(8)/7*2pi$

$= (\frac{16 π}{7}) r a d s^{- 2}$ $=((16pi)/7) rads^(-2)$

So the torque is $τ = 49 \cdot \frac{16 π}{7} N m = \frac{112}{7} π N m = 50.3 N m$ $tau=49*(16pi)/7 Nm=112/7piNm=50.3Nm$

What torque would have to be applied to a rod with a length of 7 m7m7 m and a mass of 12 kg12kg12 kg to change its horizontal spin by a frequency 8 Hz8Hz8 Hz over 7 s7s7 s?