Question

question 15
problem reference 8 - 3
a wheel of diameter 40.0 cm starts from rest and rotates with a constant angular acceleration of 3.00 rad/s². the wheel completes two revolutions.
what is the centripetal acceleration of a point on the rim of the wheel at the instant when the wheel completes its second revolution?
30.1 m/s²
3.01×10⁻³ m/s²
15.1 m/s²
21.7 m/s²

Explanation:

Step1: Convert diameter to radius

Radius $r = \frac{40.0\,\text{cm}}{2} = 20.0\,\text{cm} = 0.200\,\text{m}$

Step2: Calculate total angular displacement

Two revolutions: $\theta = 2 \times 2\pi = 4\pi\,\text{rad}$

Step3: Find final angular velocity

Use $\omega^2 = \omega_0^2 + 2\alpha\theta$ (starts from rest, $\omega_0=0$):
$\omega^2 = 0 + 2 \times 3.00\,\text{rad/s}^2 \times 4\pi\,\text{rad} = 24\pi\,\text{rad}^2/\text{s}^2$

Step4: Compute centripetal acceleration

$a_c = r\omega^2 = 0.200\,\text{m} \times 24\pi\,\text{rad}^2/\text{s}^2 \approx 0.200 \times 75.398 \approx 15.1\,\text{m/s}^2$

Answer:

15.1 m/s²