What torque would have to be applied to a rod with a length of $4 m$ $4 m$ and a mass of $3 k g$ $3 kg$ to change its horizontal spin by a frequency of $3 H z$ $3 Hz$ over $2 s$ $2 s$ ?

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What torque would have to be applied to a rod with a length of $4 m$ $4 m$ and a mass of $3 k g$ $3 kg$ to change its horizontal spin by a frequency of $3 H z$ $3 Hz$ over $2 s$ $2 s$ ?

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Answer 1 · 2016-11-18T05:42:33

The torque applied to the rod is 37.68 Newton-metre.

Explanation:

Torque is defined as the product of moment of inertia and angular acceleration
$\Rightarrow τ = I α$ $=>tau=Ialpha$
Moment of inertia of rod is $I_{rod} = (\frac{M L^{2}}{12})$ $I_"rod"=((ML^2)/12)$ and angular acceleration is change in angular velocity with respect to time
$\Rightarrow α = \frac{ω}{t}$ $=>alpha=omega/t$
angular velocity $ω = 2 π ν$ $omega=2pinu$ as frequency is given in above case.
Now the torque becomes $τ = [\frac{M L^{2}}{12}] [\frac{2 π ν}{t}]$ $tau=[(ML^2)/12][(2pinu)/t]$
Substituting the given values we get,
$τ = [\frac{(3 k g) {(4 m)}^{2}}{12}] [\frac{2 \times 3.14 \times 3 H z}{2 s}]$ $tau=[[(3kg)(4m)^2]/12][(2xx3.14xx3Hz)/(2s)]$
$:.$ torque $tau=37.68$ N-m.

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What torque would have to be applied to a rod with a length of $4 m$ $4 m$ and a mass of $3 k g$ $3 kg$ to change its horizontal spin by a frequency of $3 H z$ $3 Hz$ over $2 s$ $2 s$ ?

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Explanation:

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What torque would have to be applied to a rod with a length of $4 m$ $4 m$ and a mass of $3 k g$ $3 kg$ to change its horizontal spin by a frequency of $3 H z$ $3 Hz$ over $2 s$ $2 s$ ?