Set 8 Problem number 17

• Important Note on Notation

• Problem

• Solution

• Generalized Solution

Problem

If a disk has moment of inertia .1578554 kilogram meter ^ 2 and must be accelerated from .81 radians/second to 4.8 radians/second while rotating through 70.9 radians, what net torque must be used?

Solution

From the given information we can determine that the angular acceleration must be .1578554 radians/second ^ 2.

• The net torque required to accelerate the .1578554 kg m ^ 2 moment of inertia at this rate is `tau = I `alpha = ( .1578554 kg m ^ 2)( .1578554 radians/second ^ 2) = 2.491834E-02 meter Newtons.

Generalized Solution

Given initial and final angular velocities `omega0 and `omegaf and angular displacement `d`theta, we find the angular acceleration by the following reasoning:

• average angular velocity = ( `omega0 + `omegaf ) / 2
• time interval = angular displacement / average angular velocity

= `d`theta /[  (`omega0 + `omegaf) / 2 ]

= 2 `d`theta / (`omega0 + `omegaf)

• angular acceleration = change in angular velocity / time interval

= ( `omegaf - `omega0 ) / [ 2 `d`theta / (`omega0 + `omegaf) ]

= 2 ( `omegaf^2 - `omega0^2 ) / `d`theta.

The same result would have been obtained from the equation `omegaf^2 = `omega0^2 + 2 `alpha `d`theta.

Given the moment of inertia I we would find the torque required from Newton's Second Law in rotational form:

• torque = `tau = I `alpha = I * 2 ( `omegaf^2 - `omega0^2 ) / `d`theta.