What torque would have to be applied to a rod with a length of 12 m12m and a mass of 6 kg6kg to change its horizontal spin by a frequency 7 Hz7Hz over 9 s9s?

1 Answer
Narad T.
Jul 11, 2017

The torque for the rod rotating about the center is =351.9Nm=351.9Nm
The torque for the rod rotating about one end is =1407.4Nm=1407.4Nm

Explanation:

The torque is the rate of change of angular momentum

tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dtτ=dLdt=d(Iω)dt=Idωdt

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

I=1/12*mL^2I=112⋅mL2

=1/12*6*12^2= 72 kgm^2=112⋅6⋅122=72kgm2

The rate of change of angular velocity is

(domega)/dt=(7)/9*2pidωdt=79⋅2π

=(14/9pi) rads^(-2)=(149π)rads−2

So the torque is tau=72*(14/9pi) Nm=112piNm=351.9Nmτ=72⋅(149π)Nm=112πNm=351.9Nm

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

I=1/3*mL^2I=13⋅mL2

=1/3*6*12^2=288kgm^2=13⋅6⋅122=288kgm2

So,

The torque is tau=288*(14/9pi)=1407.4Nmτ=288⋅(149π)=1407.4Nm

Related questions
See all questions in Torque
Impact of this question
1248 views around the world
You can reuse this answer
Creative Commons License