Question

boy is riding a bicycle at a velocity of 4 m/s. The bicycle's wheel's have a diameter of 0.8m. what is acceleration of the point on the rim of the wheel?

Answer 1

To find the acceleration of the point on the rim of the wheel of the bicycle, we need to understand a few concepts about circular motion.

1. **Linear velocity (v)**: The boy is riding the bicycle at a velocity of 4 m/s.

2. **Radius (r) of the wheel**: The diameter of the wheel is 0.8 m, so the radius is:
\[
r = \frac{0.8 \, \text{m}}{2} = 0.4 \, \text{m}
\]

3. **Angular velocity (\(\omega\))**: The relationship between linear velocity and angular velocity is:
\[
v = r \cdot \omega
\]
Rearranging for \(\omega\):
\[
\omega = \frac{v}{r} = \frac{4 \, \text{m/s}}{0.4 \, \text{m}} = 10 \, \text{rad/s}
\]

4. **Centripetal acceleration (\(a_c\))**: The centripetal acceleration of a point on the rim of the wheel can be calculated using the formula:
\[
a_c = \frac{v^2}{r}
\]
Substituting in the values:
\[
a_c = \frac{(4 \, \text{m/s})^2}{0.4 \, \text{m}} = \frac{16 \, \text{m}^2/\text{s}^2}{0.4 \, \text{m}} = 40 \, \text{m/s}^2
\]

Since the point on the rim of the wheel is in circular motion, the acceleration is entirely centripetal. Therefore, the acceleration of the point on the rim of the wheel is:

\[
\text{Acceleration} = 40 \, \text{m/s}^2
\]