What torque would have to be applied to a rod with a length of $6 m$ $6 m$ and a mass of $2 k g$ $2 kg$ to change its horizontal spin by a frequency $2 H z$ $2 Hz$ over $8 s$ $8 s$ ?

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Answer 1 · 2017-04-01T17:07:01

The torque, for the rod rotating about the center is $= 9.42 N m$ $=9.42Nm$
The torque, for the rod rotating about one end is $= 37.7 N m$ $=37.7Nm$

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

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

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

The rate of change of angular velocity is

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

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

So the torque is $τ = 6 \cdot (\frac{π}{2}) N m = 9.42 N m$ $tau=6*(pi/2) Nm=9.42Nm$

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

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

$= \frac{1}{3} \cdot 2 \cdot 6^{2} = 24 k g m^{2}$ $=1/3*2*6^2=24kgm^2$

So,

The torque is $τ = 24 \cdot (\frac{π}{2}) = 37.7 N m$ $tau=24*(pi/2)=37.7Nm$

What torque would have to be applied to a rod with a length of 6 m6m6 m and a mass of 2 kg2kg2 kg to change its horizontal spin by a frequency 2 Hz2Hz2 Hz over 8 s8s8 s?