A certain wheel has a rotational inertia of 12 kg.m^2. As it turns through 5.0 rev its angular
velocity increases from 5.0 rad/s to 6.0 rad/s. If the net torque is constant its value is ?

Question

anonymous · Answer

when torque constant  we used  that eq : $$\omega=\tau (\Delta \Theta)$$
$$\tau = (6/10\pi)$$  so, Torque =1.89
but the correct answer was  2.1 N.m !!

theeric · Answer

This link might help: http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#mi

theeric · Answer

That equation can't be correct because it can be rearranged to be$$\tau =\frac{\omega [rad/s]}{\Delta\theta [rad]}$$That would mean torque is measured in $1/s$ - also called $Hz$ or Hertz.

theeric · Answer

So $\tau = I\alpha$ could work for you (it's like $F=ma$), and $\alpha$ is angular acceleration.

Is there maybe a $v_f^2=v_i^2+2ad$ sort of equation for rotation?

theeric · Answer

According to http://bama.ua.edu/~jharrell/PH105-S03/exercises/rot_mot_eqs.htm there is! $$\omega_f^2=\omega_i^2+2\alpha\left(\Delta \theta\right)$$ And $\Large\Delta\theta=5[rev]\times \frac{2\pi [rad]}{1[rev]}=10\pi [rad]$ I did the math, and I got $2.1008452488130184321492656765172$ on the Windows calculator.

anonymous · Answer

Thank you very  much :)

theeric · Answer

You're welcome! $\Large\text{:)}$