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 $15 H z$ $15 Hz$ over $9 s$ $9 s$ ?

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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 $15 H z$ $15 Hz$ over $9 s$ $9 s$ ?

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Answer 1 · 2017-03-02T20:10:41

The torque (for the rod rotating about the center) is $= 446.8 N m$ $=446.8Nm$
The torque (for the rod rotating about one end) is $= 1782.2 N m$ $=1782.2Nm$

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

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

$= \frac{1}{12} \cdot 8 \cdot 8^{2} = \frac{128}{3} k g m^{2}$ $=1/12*8*8^2= 128/3 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{15}{9} \cdot 2 π$ $(domega)/dt=(15)/9*2pi$

$= (\frac{10}{3} π) r a d s^{- 2}$ $=(10/3pi) rads^(-2)$

So the torque is $τ = \frac{128}{3} \cdot (\frac{10}{3} π) N m = \frac{1280}{9} π N m = 446.8 N m$ $tau=128/3*(10/3pi) Nm=1280/9piNm=446.8Nm$

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

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

$= \frac{1}{3} \cdot 8 \cdot 8^{2} = \frac{512}{3} k g m^{2}$ $=1/3*8*8^2=512/3kgm^2$

So,

The torque is $τ = \frac{512}{3} \cdot (\frac{10}{3} π) = \frac{5120}{9} π = 1787.2 N m$ $tau=512/3*(10/3pi)=5120/9pi=1787.2Nm$

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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 $15 H z$ $15 Hz$ over $9 s$ $9 s$ ?

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Explanation:

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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 $15 H z$ $15 Hz$ over $9 s$ $9 s$ ?