What torque would have to be applied to a rod with a length of $8 m$ and a mass of $8 kg$ to change its horizontal spin by a frequency $1 Hz$ over $9 s$ ?

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Answer 1 · 2017-02-23T15:00:47

The torque (for the rod rotating about the center) is $=29.8Nm$
The torque (for the rod rotating about one end) is $=119.1Nm$

The moment of inertia of a rod, rotating about the center is

$I=1/12*mL^2$

$=1/12*8*8^2= 128/3 kgm^2$

The rate of change of angular velocity is

$(domega)/dt=(1)/9*2pi$

$=(2/9pi) rads^(-2)$

So the torque is $tau=128/3*(2/9pi) Nm=256/27piNm=29.8Nm$

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

$I=1/3*mL^2$

$=1/3*8*8^2=512/3kgm^2$

So,

The torque is $tau=512/3*(2/9pi)=1024/27pi=119.1Nm$

What torque would have to be applied to a rod with a length of 8 m8 m and a mass of 8 kg8 kg to change its horizontal spin by a frequency 1 Hz1 Hz over 9 s9 s?