What torque would have to be applied to a rod with a length of $4 m$ $4 m$ and a mass of $4 k g$ $4 kg$ to change its horizontal spin by a frequency $7 H z$ $7 Hz$ over $1 s$ $1 s$ ?

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Answer 1 · 2017-02-17T06:54:29

The torque is $= 234.6 N m$ $=234.6Nm$

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

$= \frac{1}{12} \cdot 4 \cdot 4^{2} = \frac{16}{3} k g m^{2}$ $=1/12*4*4^2= 16/3 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{7}{1} \cdot 2 π$ $(domega)/dt=(7)/1*2pi$

$= (14 π) r a d s^{- 2}$ $=(14pi) rads^(-2)$

So the torque is $τ = \frac{16}{3} \cdot (14 π) N m = \frac{224}{3} π N m = 234.6 N m$ $tau=16/3*(14pi) Nm=224/3piNm=234.6Nm$

What torque would have to be applied to a rod with a length of 4 m4m4 m and a mass of 4 kg4kg4 kg to change its horizontal spin by a frequency 7 Hz7Hz7 Hz over 1 s1s1 s?