What torque would have to be applied to a rod with a length of $8 m$ $8 m$ and a mass of $8 k g$ $8 kg$ to change its horizontal spin by a frequency $2 H z$ $2 Hz$ over $8 s$ $8 s$ ?

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Answer 1 · 2017-05-12T06:16:23

The torque for the rod rotating about the center is $= 67.02 N$ $=67.02N$
The torque for the rod rotating about one end is $= 268.08 N$ $=268.08N$

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

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

$= \frac{1}{12} \cdot 8 \cdot 8^{2} = 42.67 k g m^{2}$ $=1/12*8*8^2= 42.67 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{2}{8} \cdot 2 π$ $(domega)/dt=(2)/8*2pi$

$= (\frac{1}{2} π) r a d s^{- 2}$ $=(1/2pi) rads^(-2)$

So the torque is $τ = 42.67 \cdot (\frac{1}{2} π) N m = 67.02 N m$ $tau=42.67*(1/2pi) Nm=67.02Nm$

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

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

$= \frac{1}{3} \cdot 8 \cdot 8^{2} = 170.67$ $=1/3*8*8^2=170.67$

So,

The torque is $τ = 36 \cdot (\frac{1}{2} π) = 268.08 N m$ $tau=36*(1/2pi)=268.08Nm$

What torque would have to be applied to a rod with a length of 8m8 m and a mass of 8kg8 kg to change its horizontal spin by a frequency 2Hz2 Hz over 8s8 s?