What torque would have to be applied to a rod with a length of $6 m$ $6 m$ and a mass of $2 k g$ $2 kg$ to change its horizontal spin by a frequency $7 H z$ $7 Hz$ over $8 s$ $8 s$ ?

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Answer 1 · 2017-10-30T06:44:11

The torque for the rod rotating about the center is $= 33 N m$ $=33Nm$
The torque for the rod rotating about one end is $= 132 N m$ $=132Nm$

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

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

The mass of the rod is $m = 2 k g$ $m=2kg$

The length of the rod is $L = 6 m$ $L=6m$

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

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{7}{8} \cdot 2 π$ $(domega)/dt=(7)/8*2pi$

$= (\frac{7}{4} π) r a d s^{- 2}$ $=(7/4pi) rads^(-2)$

So the torque is $τ = 6 \cdot (\frac{7}{4} π) N m = \frac{21}{2} π N m = 33 N m$ $tau=6*(7/4pi) Nm=21/2piNm=33Nm$

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

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

$= \frac{1}{3} \cdot 2 \cdot 6^{2} = 24 k g m^{2}$ $=1/3*2*6^2=24kgm^2$

So,

The torque is $τ = 24 \cdot (\frac{7}{4} π) = 42 π = 132 N m$ $tau=24*(7/4pi)=42pi=132Nm$

What torque would have to be applied to a rod with a length of 6m6 m and a mass of 2kg2 kg to change its horizontal spin by a frequency 7Hz7 Hz over 8s8 s?