What torque would have to be applied to a rod with a length of $1 m$ $1 m$ and a mass of $1 k g$ $1 kg$ to change its horizontal spin by a frequency of $7 H z$ $7 Hz$ over $5 s$ $5 s$ ?

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Answer 1 · 2017-07-21T14:35:06

The torque for the rod rotating about the center is $= 0.73 N m$ $=0.73Nm$
The torque for the rod rotating about one end is $= 2.93 N m$ $=2.93Nm$

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

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

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

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{7}{5} \cdot 2 π$ $(domega)/dt=(7)/5*2pi$

$= (\frac{14}{5} π) r a d s^{- 2}$ $=(14/5pi) rads^(-2)$

So the torque is $τ = \frac{1}{12} \cdot (\frac{14}{5} π) N m = \frac{7}{30} π N m = 0.73 N m$ $tau=1/12*(14/5pi) Nm=7/30piNm=0.73Nm$

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

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

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

So,

The torque is $τ = \frac{1}{3} \cdot (\frac{14}{5} π) = \frac{14}{15} π = 2.93 N m$ $tau=1/3*(14/5pi)=14/15pi=2.93Nm$

What torque would have to be applied to a rod with a length of 1 m1m1 m and a mass of 1 kg1kg1 kg to change its horizontal spin by a frequency of 7 Hz7Hz7 Hz over 5 s5s5 s?