What torque would have to be applied to a rod with a length of $5$ $5$ and a mass of $5 k g$ $5 kg$ to change its horizontal spin by a frequency of $2 H z$ $2 Hz$ over $4 s$ $4 s$ ?

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Answer 1 · 2017-06-27T12:42:13

The torque for the rod rotating about the center is $= 32.7 N m$ $=32.7Nm$
The torque for the rod rotating about one end is $= 130.9 N m$ $=130.9Nm$

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

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

$= \frac{1}{12} \cdot 5 \cdot 5^{2} = \frac{125}{12} k g m^{2}$ $=1/12*5*5^2= 125/12 kgm^2$

The rate of change of angular velocity is

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

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

So the torque is $τ = \frac{125}{12} \cdot (π) N m = 32.7 N m$ $tau=125/12*(pi) Nm=32.7Nm$

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

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

$= \frac{1}{3} \cdot 5 \cdot 5^{2} = \frac{125}{3} k g m^{2}$ $=1/3*5*5^2=125/3kgm^2$

So,

The torque is $τ = \frac{125}{3} \cdot (π) = 130.9 N m$ $tau=125/3*(pi)=130.9Nm$

What torque would have to be applied to a rod with a length of 5 55 and a mass of 5 kg5kg5 kg to change its horizontal spin by a frequency of 2 Hz2Hz2 Hz over 4 s4s4 s?