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 $4 s$ $4 s$ ?

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Answer 1 · 2017-06-06T05:58:22

The torque for the rod rotating about the center is $= 42.4 N m$ $=42.4Nm$
The torque for the rod rotating about one end is $= 169.6 N m$ $=169.6Nm$

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, 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

$\frac{d ω}{d t} = \frac{3}{4} \cdot 2 π$ $(domega)/dt=(3)/4*2pi$

$= (\frac{3}{2} π) r a d s^{- 2}$ $=(3/2pi) rads^(-2)$

So the torque is $τ = 9 \cdot (\frac{3}{2} π) N m = \frac{27}{2} π N m = 42.4 N m$ $tau=9*(3/2pi) Nm=27/2piNm=42.4Nm$

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 $τ = 36 \cdot (\frac{3}{2} π) = 54 π = 169.6 N m$ $tau=36*(3/2pi)=54pi=169.6Nm$

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 4 s4s4 s?