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

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Answer 1 · 2017-02-21T14:05:25

The torque (rod rotating about the center) $= 65.45 N m$ $=65.45Nm$
The torque (rod rotating about one end) $= 261.8 N m$ $=261.8Nm$

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

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

$= \frac{1}{12} \cdot 5 \cdot 5^{2} = \frac{125}{12} k g m^{2}$ $=1/12*5*5^2= 125/12 kgm^2$

The rate of change of angular velocity is

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

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

So the torque is $τ = \frac{125}{12} \cdot (2 π) N m = \frac{125}{6} π N m = 65.45 N m$ $tau=125/12*(2pi) Nm=125/6piNm=65.45Nm$

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

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

$= \frac{1}{3} \cdot 5 \cdot 5^{2} = \frac{125}{3}$ $=1/3*5*5^2=125/3$

So,

The torque is $τ = \frac{125}{3} \cdot (2 π) = \frac{250}{3} π = 261.8 N m$ $tau=125/3*(2pi)=250/3pi=261.8Nm$

What torque would have to be applied to a rod with a length of 5m5 m and a mass of 5kg5 kg to change its horizontal spin by a frequency of 2Hz2 Hz over 2s2 s?