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 $5 H z$ $5 Hz$ over $2 s$ $2 s$ ?

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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 $5 H z$ $5 Hz$ over $2 s$ $2 s$ ?

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Answer 1 · 2017-12-05T15:34:22

The torque for the rod rotating about its center is $= 62.83 N m$ $=62.83Nm$
The torque for the rod rotating about one end is $= 251.33 N m$ $=251.33Nm$

Explanation:

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 mass of the rod is $m = 3 k g$ $m=3kg$

The length of the rod is $L = 4 m$ $L=4m$

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 frequency is $f = 5 H z$ $f=5 Hz$

The rate of change of angular velocity is

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

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

So the torque is $τ = 4 \cdot (5 π) N m = 20 π N m = 62.83 N m$ $tau=4*(5pi) Nm=20piNm=62.83Nm$

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

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

$= \frac{1}{3} \cdot 3 \cdot 4^{2} = 16 k g m^{2}$ $=1/3*3*4^2=16kgm^2$

So,

The torque is $τ = 16 \cdot (5 π) = 80 π = 251.33 N m$ $tau=16*(5pi)=80pi=251.33Nm$

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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 $5 H z$ $5 Hz$ over $2 s$ $2 s$ ?

1 Answer

Explanation:

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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 $5 H z$ $5 Hz$ over $2 s$ $2 s$ ?