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 $3 H z$ $3 Hz$ over $6 s$ $6 s$ ?

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Answer 1 · 2017-02-15T15:56:05

The torque is $= 13.1 N m$ $=13.1Nm$

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

$= \frac{1}{12} \cdot 2 \cdot 5^{2} = \frac{25}{6} k g m^{2}$ $=1/12*2*5^2= 25/6 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{3}{6} \cdot 2 π$ $(domega)/dt=(3)/6*2pi$

$= π r a d s^{- 2}$ $=pi rads^(-2)$

So the torque is $τ = \frac{25}{6} \cdot π N m = 13.1 N m$ $tau=25/6*pi Nm=13.1Nm$

What torque would have to be applied to a rod with a length of 5m5 m and a mass of 2kg2 kg to change its horizontal spin by a frequency of 3Hz3 Hz over 6s6 s?