What torque would have to be applied to a rod with a length of $8 m$ $8 m$ and a mass of $8 k g$ $8 kg$ to change its horizontal spin by a frequency $5 H z$ $5 Hz$ over $8 s$ $8 s$ ?

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Answer 1 · 2017-06-12T06:15:51

The torquefor the rod rotating about the center is $= 167.6 N m$ $=167.6Nm$
The torque for the rod rotating about one end is $= 670.2 N m$ $=670.2Nm$

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

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

$= \frac{1}{12} \cdot 8 \cdot 8^{2} = 42.67 k g m^{2}$ $=1/12*8*8^2= 42.67 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{5}{8} \cdot 2 π$ $(domega)/dt=(5)/8*2pi$

$= (\frac{5}{4} π) r a d s^{- 2}$ $=(5/4pi) rads^(-2)$

So the torque is $τ = 42.67 \cdot (\frac{5}{4} π) N m = 167.6 N m$ $tau=42.67*(5/4pi) Nm=167.6Nm$

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

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

$= \frac{1}{3} \cdot 8 \cdot 8^{2} = 170.67 k g m^{2}$ $=1/3*8*8^2=170.67kgm^2$

So,

The torque is $τ = 170.67 \cdot (\frac{5}{4} π) = 670.2 N m$ $tau=170.67*(5/4pi)=670.2Nm$

What torque would have to be applied to a rod with a length of 8 m8m8 m and a mass of 8 kg8kg8 kg to change its horizontal spin by a frequency 5 Hz5Hz5 Hz over 8 s8s8 s?