What torque would have to be applied to a rod with a length of $5 m$ $5 m$ and a mass of $2 k g$ $2 kg$ to change its horizontal spin by a frequency of $5 H z$ $5 Hz$ over $6 s$ $6 s$ ?

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What torque would have to be applied to a rod with a length of $5 m$ $5 m$ and a mass of $2 k g$ $2 kg$ to change its horizontal spin by a frequency of $5 H z$ $5 Hz$ over $6 s$ $6 s$ ?

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Answer 1 · 2017-03-23T07:25:08

The torque, for the rod rotating about the center is $= 21.82 N m$ $=21.82Nm$
The torque, for the rod rotating about one end is $= 87.28 N m$ $=87.28Nm$

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

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

$= \frac{1}{12} \cdot 2 \cdot 5^{2} = \frac{25}{6} k g m^{2}$ $=1/12*2*5^2= 25/6 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{5}{6} \cdot 2 π$ $(domega)/dt=(5)/6*2pi$

$= (\frac{5}{3} π) r a d s^{- 2}$ $=(5/3pi) rads^(-2)$

So the torque is $τ = \frac{25}{6} \cdot (\frac{5}{3} π) N m = \frac{125}{18} π N m = 21.82 N m$ $tau=25/6*(5/3pi) Nm=125/18piNm=21.82Nm$

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

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

$= \frac{1}{3} \cdot 2 \cdot 5^{2} = \frac{50}{3}$ $=1/3*2*5^2=50/3$

So,

The torque is $τ = \frac{50}{3} (\frac{5}{3} π) = \frac{250}{9} π = 87.28 N m$ $tau=50/3(5/3pi)=250/9pi=87.28Nm$

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What torque would have to be applied to a rod with a length of $5 m$ $5 m$ and a mass of $2 k g$ $2 kg$ to change its horizontal spin by a frequency of $5 H z$ $5 Hz$ over $6 s$ $6 s$ ?

1 Answer

Explanation:

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Science

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Humanities

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What torque would have to be applied to a rod with a length of 5m5 m and a mass of 2kg2 kg to change its horizontal spin by a frequency of 5Hz5 Hz over 6s6 s?

1 Answer

Explanation:

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Impact of this question

What torque would have to be applied to a rod with a length of $5 m$ $5 m$ and a mass of $2 k g$ $2 kg$ to change its horizontal spin by a frequency of $5 H z$ $5 Hz$ over $6 s$ $6 s$ ?