An automobile tyre has a radius of 0.330 m, and its center moves forward with a linear speed of `v= 15.0 m//s`. Relative to the axle, what is the tangential speed of a point located 0.175 m from the axle?

Options

A. `1.04 m//s`
B. `2.50 m//s`
C. `4.39 m//s`
D. `7.96 m//s`

Answer

To solve the problem, we need to find the tangential speed of a point located 0.175 m from the axle of a tire that has a radius of 0.330 m and is moving forward with a linear speed of 15.0 m/s.

### Step-by-step Solution:

1. **Identify the Parameters**:
- Radius of the tire (R) = 0.330 m
- Linear speed of the center of the tire (v) = 15.0 m/s
- Distance from the axle to the point of interest (d) = 0.175 m

2. **Understand the Motion**:
The center of the tire moves forward with a speed of 15.0 m/s. The tire rolls without slipping, which means that the point of contact with the ground has a velocity of 0 m/s relative to the ground.

3. **Calculate the Angular Velocity (ω)**:
The relationship between linear speed (v) and angular velocity (ω) is given by:
\[ v = \omega R \]
Rearranging this gives:
\[ \omega = \frac{v}{R} = \frac{15.0 \, \text{m/s}}{0.330 \, \text{m}} \approx 45.45 \, \text{rad/s} \]

4. **Calculate the Tangential Speed of the Point**:
The tangential speed (v_p) of a point at a distance d from the axle is given by:
\[ v_p = \omega d \]
Substituting the values we have:
\[ v_p = 45.45 \, \text{rad/s} \times 0.175 \, \text{m} \approx 7.95 \, \text{m/s} \]

5. **Determine the Direction**:
Since the point is located below the axle and the tire rolls forward, the tangential speed of this point will be directed backward relative to the forward motion of the tire.

### Final Result:
The tangential speed of the point located 0.175 m from the axle is approximately **7.95 m/s** directed backward.