What torque would have to be applied to a rod with a length of $12 m$ $12 m$ and a mass of $7 k g$ $7 kg$ to change its horizontal spin by a frequency $5 H z$ $5 Hz$ over $4 s$ $4 s$ ?

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What torque would have to be applied to a rod with a length of $12 m$ $12 m$ and a mass of $7 k g$ $7 kg$ to change its horizontal spin by a frequency $5 H z$ $5 Hz$ over $4 s$ $4 s$ ?

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Answer 1 · 2017-04-12T06:24:01

The torque for the rod rotating about the center is $= 659.7 N m$ $=659.7Nm$
The torque for the rod rotating about one end is $= 2638.9 N m$ $=2638.9Nm$

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 inertia of a rod, rotating about the center is

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

$= \frac{1}{12} \cdot 7 \cdot 12^{2} = 84 k g m^{2}$ $=1/12*7*12^2= 84 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{5}{4} \cdot 2 π$ $(domega)/dt=(5)/4*2pi$

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

So the torque is $τ = 84 \cdot (\frac{5}{2} π) N m = 659.7 N m$ $tau=84*(5/2pi) Nm=659.7Nm$

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

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

$= \frac{1}{3} \cdot 7 \cdot 12^{2} = 336 k g m^{2}$ $=1/3*7*12^2=336 kgm^2$

So,

The torque is $τ = 336 \cdot (\frac{5}{2} π) = 2638.9 N m$ $tau=336*(5/2pi)=2638.9Nm$

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What torque would have to be applied to a rod with a length of $12 m$ $12 m$ and a mass of $7 k g$ $7 kg$ to change its horizontal spin by a frequency $5 H z$ $5 Hz$ over $4 s$ $4 s$ ?

1 Answer

Explanation:

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What torque would have to be applied to a rod with a length of 12m12 m and a mass of 7kg7 kg to change its horizontal spin by a frequency 5Hz5 Hz over 4s4 s?

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What torque would have to be applied to a rod with a length of $12 m$ $12 m$ and a mass of $7 k g$ $7 kg$ to change its horizontal spin by a frequency $5 H z$ $5 Hz$ over $4 s$ $4 s$ ?