To solve the problem of determining how fast a point on the rim of a wheel is rising given the radius, angular velocity, and angle above the horizontal, we can follow these steps:
1. Identify the given values:
- Radius of the wheel, [tex]\( r = 3 \)[/tex] meters
- Angular velocity, [tex]\( \omega = 18 \)[/tex] radians per second
- Angle above the horizontal, [tex]\( \theta = \frac{\pi}{3} \)[/tex] radians
2. Determine the linear velocity at the rim of the wheel:
- The linear velocity [tex]\( v \)[/tex] of a point on the rim of the wheel is given by the product of the radius and the angular velocity:
[tex]\[
v = r \cdot \omega
\][/tex]
- Plugging in the given values:
[tex]\[
v = 3 \, \text{m} \cdot 18 \, \text{rad/s} = 54 \, \text{m/s}
\][/tex]
3. Calculate the vertical component of the linear velocity:
- The vertical component of the linear velocity is the product of the linear velocity and the sine of the angle [tex]\( \theta \)[/tex]:
[tex]\[
v_y = v \cdot \sin(\theta)
\][/tex]
- Using [tex]\( \theta = \frac{\pi}{3} \)[/tex], which has a sine value of [tex]\( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \)[/tex]:
[tex]\[
v_y = 54 \, \text{m/s} \cdot \sin\left(\frac{\pi}{3}\right)
\][/tex]
- Simplifying further:
[tex]\[
v_y = 54 \, \text{m/s} \cdot \frac{\sqrt{3}}{2}
\][/tex]
- Calculating the numerical value, it results in:
[tex]\[
v_y \approx 46.8 \, \text{m/s}
\][/tex]
Final Answer:
The point on the rim of the wheel is rising at approximately [tex]\( 46.8 \)[/tex] meters per second when it is [tex]\( \frac{\pi}{3} \)[/tex] radians above the horizontal.
