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 4 Hz $o v e r$ $over$ 8 s#?

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Answer 1 · 2017-03-27T12:27:42

The torque, for the rod rotating about the center is $= 13.1 N m$ $=13.1Nm$
The torque, for the rod rotating about one end is $= 52.4 N m$ $=52.4Nm$

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

$\frac{d ω}{d t} = \frac{4}{8} \cdot 2 π$ $(domega)/dt=(4)/8*2pi$

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

So the torque is $τ = \frac{25}{6} \cdot (π) N m = 13.1 N m$ $tau=25/6*(pi) Nm=13.1Nm$

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

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

$= \frac{1}{3} \cdot 2 \cdot 5^{2} = \frac{50}{3} k g m^{2}$ $=1/3*2*5^2=50/3kgm^2$

So,

The torque is $τ = \frac{50}{3} \cdot (π) = 52.4 N m$ $tau=50/3*(pi)=52.4Nm$

What torque would have to be applied to a rod with a length of 5 m5m5 m and a mass of 2 kg2kg2 kg to change its horizontal spin by a frequency 4 Hz over over over 8 s#?