What torque would have to be applied to a rod with a length of $5 m$ $5 m$ and a mass of $12 k g$ $12 kg$ to change its horizontal spin by a frequency $8 H z$ $8 Hz$ over $7 s$ $7 s$ ?

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What torque would have to be applied to a rod with a length of $5 m$ $5 m$ and a mass of $12 k g$ $12 kg$ to change its horizontal spin by a frequency $8 H z$ $8 Hz$ over $7 s$ $7 s$ ?

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Answer 1 · 2017-10-30T10:34:45

The torque for the rod rotating about the center is $= 179.5 N m$ $=179.5Nm$
The torque for the rod rotating about one end is $= 718.1 N m$ $=718.1Nm$

Explanation:

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 $m = 12 k g$ $m=12kg$

The length of the rod is $L = 5 m$ $L=5m$

$= \frac{1}{12} \cdot 12 \cdot 5^{2} = 25 k g m^{2}$ $=1/12*12*5^2= 25 kgm^2$

The rate of change of angular velocity is

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

$= (\frac{16}{7} π) r a d s^{- 2}$ $=(16/7pi) rads^(-2)$

So the torque is $τ = 25 \cdot (\frac{16}{7} π) N m = \frac{400}{7} π N m = 179.5 N m$ $tau=25*(16/7pi) Nm=400/7piNm=179.5Nm$

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

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

$= \frac{1}{3} \cdot 12 \cdot 5^{2} = 100 k g m^{2}$ $=1/3*12*5^2=100kgm^2$

So,

The torque is $τ = 100 \cdot (\frac{16}{7} π) = \frac{1600}{7} π = 718.1 N m$ $tau=100*(16/7pi)=1600/7pi=718.1Nm$

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What torque would have to be applied to a rod with a length of $5 m$ $5 m$ and a mass of $12 k g$ $12 kg$ to change its horizontal spin by a frequency $8 H z$ $8 Hz$ over $7 s$ $7 s$ ?

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Explanation:

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What torque would have to be applied to a rod with a length of $5 m$ $5 m$ and a mass of $12 k g$ $12 kg$ to change its horizontal spin by a frequency $8 H z$ $8 Hz$ over $7 s$ $7 s$ ?