What torque would have to be applied to a rod with a length of $7 m$ $7 m$ and a mass of $9 k g$ $9 kg$ to change its horizontal spin by a frequency $9 H z$ $9 Hz$ over $18 s$ $18 s$ ?

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What torque would have to be applied to a rod with a length of $7 m$ $7 m$ and a mass of $9 k g$ $9 kg$ to change its horizontal spin by a frequency $9 H z$ $9 Hz$ over $18 s$ $18 s$ ?

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Answer 1 · 2017-04-05T06:22:12

The torque, for the rod rotating about the center is, $= 115.5 N m$ $=115.5Nm$
The torque, for the rod rotating about one end is, $= 461.8 N m$ $=461.8Nm$

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

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

$= \frac{1}{12} \cdot 9 \cdot 7^{2} = 36.75 k g m^{2}$ $=1/12*9*7^2= 36.75 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{9}{18} \cdot 2 π$ $(domega)/dt=(9)/18*2pi$

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

So the torque is $τ = 36.75 \cdot (π) N m = 36.75 π N m = 115.5 N m$ $tau=36.75*(pi) Nm=36.75piNm=115.5Nm$

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

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

$= \frac{1}{3} \cdot 9 \cdot 7^{2} = 147 k g m^{2}$ $=1/3*9*7^2=147kgm^2$

So,

The torque is $τ = 147 \cdot (π) = 461.8 N m$ $tau=147*(pi)=461.8Nm$

Answer 2 · 2017-04-05T07:35:03

$Torque: 36.75 π N . m$ $"Torque: "36.75pi" "N.m$

Explanation:

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$Let's apply Newton's second law for circular motion.$ $"Let's apply Newton's second law for circular motion."$

$F = m \cdot a For linear motion$ $F=m*a " For linear motion"$

$T = I \cdot α For circular motion$ $T=I*alpha " For circular motion"$

$Where T:Torque , I:Moment of Inertia , α : angular acceleration$ $"Where T:Torque , I:Moment of Inertia , "alpha:"angular acceleration"$

$I = \frac{1}{12} m \cdot l^{2}$ $I=1/12 m* l^2"$
$moment of inertia for a rod turning about its mass center$ $"moment of inertia for a rod turning about its mass center"$
$m = 9 k g, l = 7 m$ $m=9kg " , "l=7 m$

$I = \frac{1}{12} \cdot 9 \cdot 7^{2} = 36.75$ $I=1/12*9*7^2=36.75$

$alpha=(Delta omega)/(Delta t)$

$Delta omega=omega_2-omega_1$

$Delta omega=2pi(f_2-f_1)$

$f_2-f_1=9Hz$

$Delta omega=18pi$

$alpha=(18pi)/18$

$alpha=pi$

$T=36.75pi" "N*m$

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What torque would have to be applied to a rod with a length of $7 m$ $7 m$ and a mass of $9 k g$ $9 kg$ to change its horizontal spin by a frequency $9 H z$ $9 Hz$ over $18 s$ $18 s$ ?

2 Answers

Explanation:

Explanation:

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What torque would have to be applied to a rod with a length of 7 m7m7 m and a mass of 9 kg9kg9 kg to change its horizontal spin by a frequency 9 Hz9Hz9 Hz over 18 s18s18 s?

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Explanation:

Explanation:

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Impact of this question

What torque would have to be applied to a rod with a length of $7 m$ $7 m$ and a mass of $9 k g$ $9 kg$ to change its horizontal spin by a frequency $9 H z$ $9 Hz$ over $18 s$ $18 s$ ?