A flywheel has its angular speed increased uniformly from 15 rad/s to 60 rad/s in 80 s. If the diameter of the wheel is 2 ft, determine the magnitudes of the normal and tangential components of acceleration of a point on the rim of the wheel when t = 80 s, and the total distance the point travels during the time period.
Calculate the angular acceleration of the flywheel,
$$ \omega=\omega_{0}+\alpha \times t $$
Here,
\(\omega\) is the final angular velocity
\(\omega_{0}\) is the initial angular velocity
\(\alpha\) is the angular acceleration
Substitute \(60 \mathrm{rad} / \mathrm{s}\) for \(\omega, 15 \mathrm{rad} / \mathrm{s}\) for \(\omega_{0}\), and \(80 \mathrm{sec}\) for \(t\) in the above equation,
\(60=15+\alpha \times 80\)
\(\alpha=0.5625 \mathrm{rad} / \mathrm{s}^{2}\)
Calculate tangential acceleration of the flywheel,
\(a_{t}=\alpha r\)
Here,
\(a_{t}\) is the tangential acceleration.
\(r\) is the radius of flywheel.
Substitute \(1 \mathrm{ft}\) for \(r\) in the above equation,
\(a_{t}=0.5625 \times 1\)
\(=0.562 \mathrm{ft} / \mathrm{s}^{2}\)
Therefore the value of tangential acceleration is \(0.562 \mathrm{ft} / \mathrm{s}^{2}\)
Calculate normal acceleration of the flywheel,
$$ \begin{aligned} a_{n} &=\omega^{2} \times r \\ &=60^{2} \times 1 \\ &=3600 \mathrm{ft} / \mathrm{s}^{2} \end{aligned} $$
Therefore the normal acceleration of the given flywheel is \(3600 \mathrm{ft} / \mathrm{s}^{2}\)
Calculate the angular displacement of the flywheel during the time interval,
\(\omega^{2}=\omega_{0}^{2}+2 \times \alpha_{t} \times \theta\)
Here, \(\theta\) is the angular displacement.
Substitute the respective values in the above equation,
$$ \begin{aligned} 60^{2} &=15^{2}+2 \cdot(0.5625) \cdot \theta \\ \theta &=3000 \mathrm{rad} \end{aligned} $$
Calculate the total distance the point travels during the time period,
$$ \begin{aligned} s &=\theta \times r \\ &=3000 \times 1 \\ &=3000 \mathrm{ft} \end{aligned} $$
Therefore the total distance the point travels is \(3000 \mathrm{ft}\)