What torque would have to be applied to a rod with a length of $12 m$ $12 m$ and a mass of $6 k g$ $6 kg$ to change its horizontal spin by a frequency $7 H z$ $7 Hz$ over $3 s$ $3 s$ ?

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What torque would have to be applied to a rod with a length of $12 m$ $12 m$ and a mass of $6 k g$ $6 kg$ to change its horizontal spin by a frequency $7 H z$ $7 Hz$ over $3 s$ $3 s$ ?

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Answer 1 · 2017-05-04T12:52:20

The torque for the rod rotating about the center is $= 1055.6 N m$ $=1055.6Nm$
The torque for the rod rotating about one end is $= 4222.3 N m$ $=4222.3Nm$

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

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

$= \frac{1}{12} \cdot 6 \cdot 12^{2} = 72 k g m^{2}$ $=1/12*6*12^2= 72 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{7}{3} \cdot 2 π$ $(domega)/dt=(7)/3*2pi$

$= (\frac{14}{3} π) r a d s^{- 2}$ $=(14/3pi) rads^(-2)$

So the torque is $τ = 72 \cdot (\frac{14}{3} π) N m = 1055.6 N m$ $tau=72*(14/3pi) Nm=1055.6Nm$

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

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

$= \frac{1}{3} \cdot 6 \cdot 12^{2} = 288 k g m^{2}$ $=1/3*6*12^2=288kgm^2$

So,

The torque is $τ = 288 \cdot (\frac{14}{3} π) = 4222.3 N m$ $tau=288*(14/3pi)=4222.3Nm$

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What torque would have to be applied to a rod with a length of $12 m$ $12 m$ and a mass of $6 k g$ $6 kg$ to change its horizontal spin by a frequency $7 H z$ $7 Hz$ over $3 s$ $3 s$ ?

1 Answer

Explanation:

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What torque would have to be applied to a rod with a length of $12 m$ $12 m$ and a mass of $6 k g$ $6 kg$ to change its horizontal spin by a frequency $7 H z$ $7 Hz$ over $3 s$ $3 s$ ?