Set 8 Problem number 10

• Important Note on Notation

• Problem

• Solution

• Generalized Solution

Problem

How long does it require for an object moving around a circle to sweep out a radial angleof 5.75 radians, if during that time it increases its angular velocity from 6 radians/second to 13 radians/second?  What is its angular acceleration?

Solution

From the two angular velocities and the fact that the angular acceleration is constant we conclude that the average angular velocity is

• `omegaAve = ave angular velocity = ( 6 radians/second + 13 radians/second)/2 = 9.5 radians/second.

At this rate the time required to turn through 5.75 radians will be

• ( 5.75 radians)/( 9.5 radians/second) = .6052 radians/second ( time interval = angular displacement / average angular velocity:   `dt = `d`theta / `omegaAve )

From the two velocities we can also determine that the change in velocity is 7 radians/second.

• To accomplish this change in .6052 seconds requires an acceleration of ( 7 radians/second) / ( .6052 seconds) = 4.236 radians/s^2.

Generalized Solution

If we know

`ds, v0 and vf we can determine vAve, then divide `ds by vAve to find `dt:

• vAve = (v0 + vf) / 2
• `dt = `ds / vAve
• We can calculate the result in two steps or combine the steps to obtain `dt  = `ds / [ (v0 + vf) / 2 ] = 2 `ds / (v0 + vf).

In the present example we know `d`theta, `omega0 and `omegaf.  So we can determine `omegaAve, then divide `d`theta by `omegaAve to find `dt:

• `omegaAve = (`omega0 + `omegaf) / 2
• `dt = `d`theta / `omegaAve
• We can calculate the result in two steps or combine the steps to obtain `dt = `d`theta / [ (`omega0 + `omegaf) / 2 ] = 2 `d`theta / (`omega0 + `omegaf).

Note that the reasoning is identical in the two situations.