Problem

# A flywheel has its angular speed increased uniformly from 15 rad/s to 60 rad/s in 80 s. If...

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.

#### Step-by-Step Solution

Solution 1

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 $$3000 \mathrm{ft}$$