What torque would have to be applied to a rod with a length of $4 m$ $4 m$ and a mass of $5 k g$ $5 kg$ to change its horizontal spin by a frequency of $3 H z$ $3 Hz$ over $7 s$ $7 s$ ?

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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 $5 k g$ $5 kg$ to change its horizontal spin by a frequency of $3 H z$ $3 Hz$ over $7 s$ $7 s$ ?

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Answer 1 · 2017-05-30T06:30:04

The torque for the rod rotating about the center is $= 17.95 N m$ $=17.95Nm$
The torque for the rod rotating about one end is $= 71.81 N m$ $=71.81Nm$

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

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

$= \frac{1}{12} \cdot 5 \cdot 4^{2} = 6.67 k g m^{2}$ $=1/12*5*4^2= 6.67 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{3}{7} \cdot 2 π$ $(domega)/dt=(3)/7*2pi$

$= (\frac{6}{7} π) r a d s^{- 2}$ $=(6/7pi) rads^(-2)$

So the torque is $τ = 6.67 \cdot (\frac{6}{7} π) N m = 17.95 N m$ $tau=6.67*(6/7pi) Nm=17.95Nm$

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

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

$= \frac{1}{3} \cdot 5 \cdot 4^{2} = 26.67 k g m^{2}$ $=1/3*5*4^2=26.67kgm^2$

So,

The torque is $τ = 26.67 \cdot (\frac{6}{7} π) = 71.81 N m$ $tau=26.67*(6/7pi)=71.81Nm$

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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 $5 k g$ $5 kg$ to change its horizontal spin by a frequency of $3 H z$ $3 Hz$ over $7 s$ $7 s$ ?

1 Answer

Explanation:

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What torque would have to be applied to a rod with a length of 4m4 m and a mass of 5kg5 kg to change its horizontal spin by a frequency of 3Hz3 Hz over 7s7 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 $5 k g$ $5 kg$ to change its horizontal spin by a frequency of $3 H z$ $3 Hz$ over $7 s$ $7 s$ ?