What torque would have to be applied to a rod with a length of $4 m$ $4 m$ and a mass of $3 k g$ $3 kg$ to change its horizontal spin by a frequency of $13 H z$ $13 Hz$ over $4 s$ $4 s$ ?

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Answer 1 · 2017-12-14T06:40:53

The torque for the rod rotating about the center is $= 81.7 N m$ $=81.7Nm$
The torque for the rod rotating about one end is $= 326.7 N m$ $=326.7Nm$

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 mass of the rod is $m = 3 k g$ $m=3kg$

The length of the rod is $L = 4 m$ $L=4m$

The moment of inertia of a rod, rotating about the center is

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

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

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{13}{4} \cdot 2 π$ $(domega)/dt=(13)/4*2pi$

$= (\frac{13}{2} π) r a d s^{- 2}$ $=(13/2pi) rads^(-2)$

So the torque is $τ = 4 \cdot (\frac{13}{2} π) N m = 26 π N m = 81.7 N m$ $tau=4*(13/2pi) Nm=26piNm=81.7Nm$

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

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

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

So,

The torque is $τ = 16 \cdot (\frac{13}{2} π) = 104 π = 326.7 N m$ $tau=16*(13/2pi)=104pi=326.7Nm$

What torque would have to be applied to a rod with a length of 4m4 m and a mass of 3kg3 kg to change its horizontal spin by a frequency of 13Hz13 Hz over 4s4 s?