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 $1 H z$ $1 Hz$ over $2 s$ $2 s$ ?

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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 $1 H z$ $1 Hz$ over $2 s$ $2 s$ ?

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Answer 1 · 2017-03-29T06:36:40

The torque, for the rod rotating about the center is $= 134.04 N m$ $=134.04Nm$
The torque, for the rod rotating about one end is $= 536.17 N m$ $=536.17Nm$

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

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

$= \frac{1}{12} \cdot 8 \cdot 8^{2} = 42.67 k g m^{2}$ $=1/12*8*8^2=42.67 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{1}{2} \cdot 2 π$ $(domega)/dt=(1)/2*2pi$

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

So the torque is $τ = 42.67 \cdot (π) N m = 134.04 N m$ $tau=42.67*(pi) Nm=134.04Nm$

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

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

$= \frac{1}{3} \cdot 8 \cdot 8^{2} = 170.67$ $=1/3*8*8^2=170.67$

So,

The torque is $τ = 170.67 \cdot (π) = \frac{504}{5} π = 536.17 N m$ $tau=170.67*(pi)=504/5pi=536.17Nm$

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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 $1 H z$ $1 Hz$ over $2 s$ $2 s$ ?

1 Answer

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 $1 H z$ $1 Hz$ over $2 s$ $2 s$ ?