What torque would have to be applied to a rod with a length of $1 m$ $1 m$ and a mass of $2 k g$ $2 kg$ to change its horizontal spin by a frequency of $3 H z$ $3 Hz$ over $3 s$ $3 s$ ?

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Answer 1 · 2016-12-23T14:35:03

The torque is $= 1.05 N m$ $=1.05Nm$

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 is $I = \frac{1}{12} m l^{2}$ $I=1/12ml^2$

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

The rate of change of angular velocity is

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

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

So the torque is $τ = \frac{1}{6} \cdot 2 π N m = \frac{π}{3} N m = 1.05 N m$ $tau=1/6*2pi Nm=pi/3Nm=1.05Nm$

What torque would have to be applied to a rod with a length of 1m1 m and a mass of 2kg2 kg to change its horizontal spin by a frequency of 3Hz3 Hz over 3s3 s?