What torque would have to be applied to a rod with a length of $3 m$ $3 m$ and a mass of $8 k g$ $8 kg$ to change its horizontal spin by a frequency $8 H z$ $8 Hz$ over $5 s$ $5 s$ ?

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Answer 1 · 2017-06-05T05:55:15

The torque for the rod rotating about the center is $= 60.3 N m$ $=60.3Nm$
The torque for the rod rotating about one end is $= 241.3 N m$ $=241.3Nm$

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

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

$= \frac{1}{12} \cdot 8 \cdot 3^{2} = 6 k g m^{2}$ $=1/12*8*3^2= 6 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{8}{5} \cdot 2 π$ $(domega)/dt=(8)/5*2pi$

$= (\frac{16}{5} π) r a d s^{- 2}$ $=(16/5pi) rads^(-2)$

So the torque is $τ = 6 \cdot (\frac{16}{5} π) N m = \frac{96}{5} π N m = 60.3 N m$ $tau=6*(16/5pi) Nm=96/5piNm=60.3Nm$

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

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

$= \frac{1}{3} \cdot 8 \cdot 3^{2} = 24 k g m^{2}$ $=1/3*8*3^2=24kgm^2$

So,

The torque is $τ = 24 \cdot (\frac{16}{5} π) = \frac{3844}{5} π = 241.3 N m$ $tau=24*(16/5pi)=3844/5pi=241.3Nm$

What torque would have to be applied to a rod with a length of 3m3 m and a mass of 8kg8 kg to change its horizontal spin by a frequency 8Hz8 Hz over 5s5 s?