What torque would have to be applied to a rod with a length of $3 m$ $3 m$ and a mass of $4 k g$ $4 kg$ to change its horizontal spin by a frequency $7 H z$ $7 Hz$ over $1 s$ $1 s$ ?

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What torque would have to be applied to a rod with a length of $3 m$ $3 m$ and a mass of $4 k g$ $4 kg$ to change its horizontal spin by a frequency $7 H z$ $7 Hz$ over $1 s$ $1 s$ ?

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Answer 1 · 2017-02-26T08:36:40

The torque (for the rod rotating about the center) is $= 131.95 N m$ $=131.95Nm$
The torque (for the rod rotating about one end) is $= 527.8 N m$ $=527.8Nm$

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 moment of inertiaof a rod, rotating about the center is

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

$= \frac{1}{12} \cdot 4 \cdot 3^{2} = 3 k g m^{2}$ $=1/12*4*3^2= 3 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{7}{1} \cdot 2 π$ $(domega)/dt=(7)/1*2pi$

$= (14 π) r a d s^{- 2}$ $=(14pi) rads^(-2)$

So the torque is $τ = 3 \cdot (14 π) N m = 42 π N m = 131.95 N m$ $tau=3*(14pi) Nm=42piNm=131.95Nm$

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

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

$= \frac{1}{3} \cdot 4 \cdot 3^{2} = 12 k g m^{2}$ $=1/3*4*3^2=12kgm^2$

So,

The torque is $τ = 12 \cdot (14 π) = 527.8 N m$ $tau=12*(14pi)=527.8Nm$

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What torque would have to be applied to a rod with a length of $3 m$ $3 m$ and a mass of $4 k g$ $4 kg$ to change its horizontal spin by a frequency $7 H z$ $7 Hz$ over $1 s$ $1 s$ ?

1 Answer

Explanation:

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What torque would have to be applied to a rod with a length of $3 m$ $3 m$ and a mass of $4 k g$ $4 kg$ to change its horizontal spin by a frequency $7 H z$ $7 Hz$ over $1 s$ $1 s$ ?