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 $9 H z$ $9 Hz$ over $3 s$ $3 s$ ?

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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 $9 H z$ $9 Hz$ over $3 s$ $3 s$ ?

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Answer 1 · 2018-01-27T18:21:00

Torque ( $τ$ $tau$ ) is defined as change in rate of angular momentum ( $L$ $L$ )i.e $\frac{d L}{d t}$ $(dL)/dt$ i.e $\frac{d (I ω)}{d t}$ $(d(Iomega))/dt$ i.e $I \frac{d ω}{d t}$ $I (domega)/dt$ (where, $I$ $I$ nis the moment of inertia)

Now,moment of inertia of a rod w.r.t its one end along an axis perpendicular to its plane is $\frac{M L^{2}}{3}$ $(ML^2)/3$ i.e $2 \cdot \frac{{(5)}^{2}}{3}$ $2*(5)^2/3$ or $\frac{50}{3} K g . m^{2}$ $50/3 Kg.m^2$

Now, as the radius of the circle isn't mentioned,so the rod can move about its mid point as well,in that case $I = \frac{M L^{2}}{12}$ $I = (ML^2)/12$ i.e $\frac{25}{6} K g . m^{2}$ $25/6 Kg.m^2$

Now, rate of change in angular velocity = $\frac{d ω}{d t}$ $(domega)/dt$ = $2 π \frac{ν 1 - ν 2}{t}$ $2pi (nu1-nu2)/t$ i.e $2 π \cdot \frac{9}{3}$ $2 pi*9/3$ or $6 π$ $6 pi$

So, $τ = \frac{50}{3} \cdot 6 π$ $tau = 50/3* 6 pi$ or, $314.16 N . m$ $314.16 N.m$ or, $\frac{25}{6} \cdot 6 π$ $25/6*6pi$ i.e $78.54 N . m$ $78.54 N.m$ for the $2 n d$ $2nd$ case

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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 $9 H z$ $9 Hz$ over $3 s$ $3 s$ ?

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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 $9 H z$ $9 Hz$ over $3 s$ $3 s$ ?