Set 8 Problem number 11

• Important Note on Notation

• Problem

• Solution

• Generalized Solution

Problem

A rolling ball accelerates uniformly at .3 radians/second ^ 2. At a certain instant its angular velocity is 2 radians/second.

• From this instant, how long does it require to then rotate through 11 radians, and what angular velocity does it attain in the process?

Solution

There is no simple way to reason this problem out. So we use the analog to the equation of uniformly accelerating linear motion which relates vf, v0, a and ds: vf ^ 2 = v0 ^ 2 + 2a (ds).

• The equation would be expressed as `omegaf^2 =`omega0^2 + 2 `alpha `dTheta.
• We obtain final angular velocity

• `omegaf = `sqrt(`omega0^2 + 2 `alpha `dTheta) = +- `sqrt [( 2 rad/sec) ^ 2 + 2( .3 rad/sec ^ 2)( 11 rad)] = +- 3.255764 rad/sec.

• We can reject the negative result, which is meaningless in this context, and we conclude that the final angular velocity is 3.255764 rad/sec.

From the initial and final angular velocities, we obtain

• average anglular velocity = 2.627882 rad/s,

from which we can calculate the time required to rotate through 11 radians.

The result is `dt = ( 11 rad)/( 2.627882 rad/sec) = 4.185881 sec.

Generalized Solution

If we know

`ds, v0 and a we can use vf^2 = v0^2 + 2 a `ds to determine vf.  We can then average vf and v0 to obtain vAve, which we use with `ds to determine `dt:

• vf = +- `sqrt( v0^2 + 2 a `ds)
• vAve = (v0 + vf) / 2
• `dt = `ds / vAve.

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

• `omegaf = +- `sqrt(`omega0^2 + 2 `alpha `d`theta)
• `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.