What torque would have to be applied to a rod with a length of $5 m$ $5 m$ and a mass of $2 k g$ $2 kg$ to change its horizontal spin by a frequency of $24 H z$ $24 Hz$ over $8 s$ $8 s$ ?

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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 $2 k g$ $2 kg$ to change its horizontal spin by a frequency of $24 H z$ $24 Hz$ over $8 s$ $8 s$ ?

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Answer 1 · 2016-02-13T22:35:35

$L = 25 π$ $L=25pi$

Explanation:

$torque is given by formula:$ $"torque is given by formula:"$
$L = α \cdot I$ $L=alpha*I$
$if rod spinning around of its center: I = \frac{1}{12} \cdot m \cdot L^{2}$ $"if rod spinning around of its center: " I=1/12*m*L^2$
$else use I = \frac{1}{3} \cdot m \cdot L^{2}$ $"else use " I=1/3*m*L^2$
$I = \frac{1}{12} \cdot 2 \cdot 5^{2}$ $I=1/12*2*5^2$
$I = \frac{1}{12} \cdot 50$ $I=1/12*50$
$I = \frac{50}{12}$ $I=50/12$
$α = 2 π \cdot \frac{24}{8} = 6 π$ $alpha=2pi*24/8=6pi$
$L = 6 π \cdot \frac{50}{12}$ $L=6pi*50/12$
$L = 25 π$ $L=25pi$

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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 $2 k g$ $2 kg$ to change its horizontal spin by a frequency of $24 H z$ $24 Hz$ over $8 s$ $8 s$ ?

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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 $2 k g$ $2 kg$ to change its horizontal spin by a frequency of $24 H z$ $24 Hz$ over $8 s$ $8 s$ ?