What torque would have to be applied to a rod with a length of $1 m$ $1 m$ and a mass of $2 k g$ $2 kg$ to change its horizontal spin by a frequency of $9 H z$ $9 Hz$ over $4 s$ $4 s$ ?

Question

1434 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-03T15:19:55

The torque for the rod rotating about the center is $= 2.36 N m$ $=2.36Nm$
The torque for the rod rotating about one end is $= 9.42 N m$ $=9.42Nm$

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

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

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

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{9}{4} \cdot 2 π$ $(domega)/dt=(9)/4*2pi$

$= (\frac{9}{2} π) r a d s^{- 2}$ $=(9/2pi) rads^(-2)$

So the torque is $τ = \frac{1}{6} \cdot (\frac{9}{2} π) N m = \frac{3}{4} π N m = 2.36 N m$ $tau=1/6*(9/2pi) Nm=3/4piNm=2.36Nm$

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

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

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

So,

The torque is $τ = \frac{2}{3} \cdot (\frac{9}{2} π) = 3 π = 9.42 N m$ $tau=2/3*(9/2pi)=3pi=9.42Nm$

What torque would have to be applied to a rod with a length of 1 m1m1 m and a mass of 2 kg2kg2 kg to change its horizontal spin by a frequency of 9 Hz9Hz9 Hz over 4 s4s4 s?