What torque would have to be applied to a rod with a length of $7 m$ $7 m$ and a mass of $9 k g$ $9 kg$ to change its horizontal spin by a frequency $7 H z$ $7 Hz$ over $18 s$ $18 s$ ?

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Answer 1 · 2017-07-08T09:33:50

The torque for the rod rotating about the center is $= 89.8 N m$ $=89.8Nm$
The torque for the rod rotating about one end is $= 359.2 N m$ $=359.2Nm$

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, rotating about the center is

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

$= \frac{1}{12} \cdot 9 \cdot 7^{2} = \frac{147}{4} k g m^{2}$ $=1/12*9*7^2= 147/4 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{7}{18} \cdot 2 π$ $(domega)/dt=(7)/18*2pi$

$= (\frac{7}{9} π) r a d s^{- 2}$ $=(7/9pi) rads^(-2)$

So the torque is $τ = \frac{147}{4} \cdot (\frac{7}{9} π) N m = 89.8 N m$ $tau=147/4*(7/9pi) Nm=89.8Nm$

The moment of inertia of a rod, rotating about one end is

$I = \frac{1}{3} \cdot m L^{2}$ $I=1/3*mL^2$

$= \frac{1}{3} \cdot 9 \cdot 7^{2} = 147 k g m^{2}$ $=1/3*9*7^2=147kgm^2$

So,

The torque is $τ = 147 \cdot (\frac{7}{9} π) = 359.2 N m$ $tau=147*(7/9pi)=359.2Nm$

What torque would have to be applied to a rod with a length of 7m7 m and a mass of 9kg9 kg to change its horizontal spin by a frequency 7Hz7 Hz over 18s18 s?