What torque would have to be applied to a rod with a length of $7 m$ $7 m$ and a mass of $9 k g$ $9 kg$ to change its horizontal spin by a frequency $12 H z$ $12 Hz$ over $16 s$ $16 s$ ?

Question

1297 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-21T07:48:25

The torque (about the center) is $= 173.2 N m$ $=173.2Nm$
The torque (about one end) is $= 692.7 N m$ $=692.7Nm$

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 abour the center is

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

$= \frac{1}{12} \cdot 9 \cdot 7^{2} = \frac{147}{4} k g m^{2}$ $=1/12*9*7^2= 147/4 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{12}{16} \cdot 2 π$ $(domega)/dt=(12)/16*2pi$

$= (\frac{3 π}{2}) r a d s^{- 2}$ $=((3pi)/2) rads^(-2)$

So the torque is $τ = \frac{147}{4} \cdot \frac{3 π}{2} N m = \frac{441}{8} π N m = 173.2 N m$ $tau=147/4*(3pi)/2 Nm=441/8piNm=173.2Nm$

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

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

$= \frac{1}{3} \cdot 9 \cdot 7^{2} = 147$ $=1/3*9*7^2=147$

So,

The torque is $τ = 147 \cdot (\frac{3}{2} π) = \frac{441}{2} π = 692.7 N m$ $tau=147*(3/2pi)=441/2pi=692.7Nm$

What torque would have to be applied to a rod with a length of 7m7 m and a mass of 9kg9 kg to change its horizontal spin by a frequency 12Hz12 Hz over 16s16 s?