58. A roller in a printing press turns through an angle \({\bf{\theta }}\left( {\bf{t}} \right)\) given by \(\theta \left( t \right) = \gamma {t^2} - \beta {t^3}\), where \(\gamma = {\bf{3}}{\bf{.2}}{\rm{0 }}{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}\), and \(\beta = 0.500{\rm{ }}{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {{{\rm{s}}^{\rm{3}}}}}} \right.} {{{\rm{s}}^{\rm{3}}}}}\).

(a) Calculate the angular velocity of the roller as a function of time.

(b) Calculate the angular acceleration of the roller as a function of time.

(c) what is the maximum positive angular velocity, and at what value of t does it occur?

The instantaneous angular velocity of a particle moving in a circular path or of a rigid object spinning around a fixed axis.

\(\omega \equiv \frac{{d\theta }}{{dt}}\) ….. (1)

The instantaneous angular acceleration of a particle moves in a circular path or of a rigid object rotating around a fixed axis.

\(\alpha \equiv \frac{{d\omega }}{{dt}}\) ….. (2)

If a rigid object rotates around a fixed axis, all part of the object has the same angular and the same angular acceleration.

Based on equation (1), the angular velocity as a function of time would be found by differentiating the angular position equation with respect to time:

\(\begin{aligned}{}\omega \left( t \right) = \frac{{d\theta \left( t \right)}}{{dt}}\\ = \frac{d}{{dt}}\left( {\gamma {t^2} - \beta {t^2}} \right)\\ = 2\gamma t - 3\beta {t^2}\end{aligned}\)

Put the numerical values for \(\gamma \) and \(\beta \) as follows:

\(\begin{aligned}\omega \left( t \right) = 2\left( {3.20{{rad} \mathord{\left/ {\vphantom {{rad} {{s^2}}}} \right. } {{s^2}}}} \right)t - 3\left( {0.500{{rad} \mathord{\left/ {\vphantom {{rad} {{s^3}}}} \right.} {{s^3}}}} \right){t^2}\\ = \left( {6.40{{rad} \mathord{\left/ {\vphantom {{rad} {{s^2}}}} \right.} {{s^2}}}} \right)t - \left( {1.50{{rad} \mathord{\left/ {\vphantom {{rad} {{s^3}}}} \right.} {{s^3}}}} \right){t^2}\,\,\,...\left( 3 \right)\end{aligned}\)

Thus, the angular velocity of the roller as a function of time is \(\omega \left( t \right) = \left( {6.40{{rad} \mathord{\left/ {\vphantom {{rad} {{s^2}}}} \right.} {{s^2}}}} \right)t - \left( {1.50{{rad} \mathord{\left/ {\vphantom {{rad} {{s^3}}}} \right.} {{s^3}}}} \right){t^2}\).

Based on equation (2), the angular acceleration as a function of time would be found by differentiating the angular velocity equation with respect to time:

\(\begin{aligned}\alpha \left( t \right) = \frac{{d\omega \left( t \right)}}{{dt}}\\ = \frac{d}{{dt}}\left( {2\gamma r - 3\beta {t^2}} \right)\\ = 2\gamma - 6\beta t\end{aligned}\)

Put the numerical values of \(\gamma \) and \(\beta \) as follows:

\(\begin{aligned}\alpha \left( t \right) = 2\left( {3.{\rm{2 }}{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right) - 6\left( {0.5{\rm{00 }}{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {{{\rm{s}}^{\rm{3}}}}}} \right.} {{{\rm{s}}^{\rm{3}}}}}} \right)t\\ =6.4{\rm{0 }}{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}} - \left( {3.0{\rm{0 }}{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {{{\rm{s}}^{\rm{3}}}}}} \right.} {{{\rm{s}}^{\rm{3}}}}}} \right)t\end{aligned}\)

Thus, the angular acceleration of the roller as a function of time is \(\alpha \left( t \right) = {\rm{6}}{\rm{.40 }}{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}} - \left( {3.0{\rm{0 }}{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}}{{{\rm{s}}^{\rm{3}}}}}} \right. } {{{\rm{s}}^{\rm{3}}}}}} \right)t\).

The maximum angular velocity happens for \(\alpha \left( t \right) = 0\). As a result,

\(\begin{aligned}\alpha \left( t \right) = 0\\6.{\rm{40}}{{{\rm{ rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {{{\rm{s}}^{\rm{2}}}}}} \right. } {{{\rm{s}}^{\rm{2}}}}} - \left( {3.{\rm{00 }}{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {{{\rm{s}}^{\rm{3}}}}}} \right. } {{{\rm{s}}^{\rm{3}}}}}} \right)t = 0\\\left( {3.0{\rm{0 }}{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {{{\rm{s}}^{\rm{3}}}}}} \right.} {{{\rm{s}}^{\rm{3}}}}}} \right)t = 6.{\rm{40 }}{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}\end{aligned}\)\(\begin{aligned}{}t = \frac{{{\rm{6}}{\rm{.40}}{{{\rm{ rad}}} \mathord{\left/ {\vphantom {{{\rm{ rad}}} {{{\rm{s}}^{\rm{2}}}}}} \right.}{{{\rm{s}}^{\rm{2}}}}}}}{{{\rm{3}}{\rm{.00 }}{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}}{{{\rm{s}}^{\rm{3}}}}}} \right.} {{{\rm{s}}^{\rm{3}}}}}}}\\ = 2.{\rm{13 s}}\end{aligned}\)

This would be the time in which the angular velocity would be maximum.

Put for this \(t\) in equation (3) to determine the angular velocity as follows:

\(\begin{aligned}{\omega _{\max }} = \left( {6.{\rm{40 }}{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\left( {2.1{\rm{3 s}}} \right) - \left({1.5{\rm{0 }}{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {{{\rm{s}}^{\rm{3}}}}}} \right.}{{{\rm{s}}^{\rm{3}}}}}} \right){\left( {{\rm{2}}{\rm{.13 s}}} \right)^2}\\ = 6.8{\rm{3 }}{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

Thus, the maximum angular velocity is \({\omega _{\max }} = {\rm{6}}{\rm{.83 }}{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\) at \(t = {\rm{2}}{\rm{.13 s}}\).