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 $2 H z$ $2 Hz$ over $1 s$ $1 s$ ?

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Answer 1 · 2017-01-03T16:05:10

The torque is $= 50.3 N m$ $=50.3Nm$

We asume that the rod is spinning around its centerm this is important for the moment of inertia.

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

$= \frac{1}{12} \cdot 3 \cdot 4^{2} = 4 k g m^{2}$ $=1/12*3*4^2= 4 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{2}{1} \cdot 2 π$ $(domega)/dt=(2)/1*2pi$

$= (4 π) r a d s^{- 2}$ $=(4pi) rads^(-2)$

So the torque is $τ = 4 \cdot 4 π N m = 16 π N m = 50.3 N m$ $tau=4*4pi Nm=16piNm=50.3Nm$

What torque would have to be applied to a rod with a length of 4m4 m and a mass of 3kg3 kg to change its horizontal spin by a frequency of 2Hz2 Hz over 1s1 s?