What torque would have to be applied to a rod with a length of $1 m$ $1 m$ and a mass of $2 k g$ $2 kg$ to change its horizontal spin by a frequency of $15 H z$ $15 Hz$ over $4 s$ $4 s$ ?

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Answer 1 · 2017-07-14T15:29:13

The torque for the rod rotating about the center is $= 3.93 N m$ $=3.93Nm$
The torque for the rod rotating about one end is $= 15.71 N m$ $=15.71Nm$

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

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

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

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{15}{4} \cdot 2 π$ $(domega)/dt=(15)/4*2pi$

$= (\frac{15}{2} π) r a d s^{- 2}$ $=(15/2pi) rads^(-2)$

So the torque is $τ = \frac{1}{6} \cdot (\frac{15}{2} π) N m = \frac{5}{4} π N m = 3.93 N m$ $tau=1/6*(15/2pi) Nm=5/4piNm=3.93Nm$

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

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

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

So,

The torque is $τ = \frac{2}{3} \cdot (\frac{15}{2} π) = 5 π = 15.71 N m$ $tau=2/3*(15/2pi)=5pi=15.71Nm$

What torque would have to be applied to a rod with a length of 1m1 m and a mass of 2kg2 kg to change its horizontal spin by a frequency of 15Hz15 Hz over 4s4 s?