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 $3 H z$ $3 Hz$ over $5 s$ $5 s$ ?

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Answer 1 · 2017-03-11T07:01:02

The torque (for rod rotating about the center ) is $= 33.9 N m$ $=33.9Nm$
The torque (for rod rotating about onr end ) is $= 135.7 N m$ $=135.7Nm$

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

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

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

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{3}{5} \cdot 2 π$ $(domega)/dt=(3)/5*2pi$

$= (\frac{6}{5} π) r a d s^{- 2}$ $=(6/5pi) rads^(-2)$

So the torque is $τ = 9 \cdot (\frac{6}{5} π) N m = \frac{54}{5} π N m = 33.9 N m$ $tau=9*(6/5pi) Nm=54/5piNm=33.9Nm$

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

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

$= \frac{1}{3} \cdot 3 \cdot 6^{2} = 36$ $=1/3*3*6^2=36$

So,

The torque is $τ = 36 \cdot (\frac{6}{5} π) = 135.7 N m$ $tau=36*(6/5pi)=135.7Nm$

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 3 Hz3Hz3 Hz over 5 s5s5 s?