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 $4 H z$ $4 Hz$ over $3 s$ $3 s$ ?

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Answer 1 · 2017-05-06T09:18:31

The torque for the rod rotating about the center is $= 33.5 N m$ $=33.5Nm$
The torque for the rod rotating about one end is $= 134.0 N m$ $=134.0Nm$

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, rotating about the center 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

$\frac{d ω}{d t} = \frac{4}{3} \cdot 2 π$ $(domega)/dt=(4)/3*2pi$

$= (\frac{8}{3} π) r a d s^{- 2}$ $=(8/3pi) rads^(-2)$

So the torque is $τ = 4 \cdot (\frac{8}{3} π) N m = \frac{32}{3} π N m = 33.5 N m$ $tau=4*(8/3pi) Nm=32/3piNm=33.5Nm$

The moment of inertia of a rod, rotating about one end is

$I = \frac{1}{3} \cdot m L^{2}$ $I=1/3*mL^2$

$= \frac{1}{3} \cdot 3 \cdot 4^{2} = 16$ $=1/3*3*4^2=16$

So,

The torque is $τ = 16 \cdot (\frac{8}{3} π) = \frac{128}{3} π = 134.0 N m$ $tau=16*(8/3pi)=128/3pi=134.0Nm$

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 4Hz4 Hz over 3s3 s?