What torque would have to be applied to a rod with a length of $8 m$ $8 m$ and a mass of $8 k g$ $8 kg$ to change its horizontal spin by a frequency $1 H z$ $1 Hz$ over $5 s$ $5 s$ ?

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Answer 1 · 2017-01-29T16:09:50

The torque is $= 53.62 N m$ $=53.62Nm$

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

$= \frac{1}{12} \cdot 8 \cdot 8^{2} = \frac{128}{3} k g m^{2}$ $=1/12*8*8^2= 128/3 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{1}{5} \cdot 2 π$ $(domega)/dt=(1)/5*2pi$

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

So the torque is $τ = \frac{128}{3} \cdot \frac{2 π}{5} N m = \frac{256}{15} π N m = 53.62 N m$ $tau=128/3*(2pi)/5 Nm=256/15piNm=53.62Nm$

What torque would have to be applied to a rod with a length of 8 m8m8 m and a mass of 8 kg8kg8 kg to change its horizontal spin by a frequency 1 Hz1Hz1 Hz over 5 s5s5 s?