A fairground ride spins its occupants inside a large teacup. If the radius of the circular path is 15.0 m15.0 m and the acceleration is 1.75 g1.75 g, what is the speed of the riders?


To find the speed of the riders on the fairground ride, we can use the formula for centripetal acceleration:

\[
a_c = \frac{v^2}{r}
\]

Where:
- \( a_c \) is the centripetal acceleration,
- \( v \) is the speed of the riders, and
- \( r \) is the radius of the circular path.

Given:
- The radius \( r = 15.0 \, \text{m} \),
- The acceleration \( a_c = 1.75 \, g \), where \( g = 9.81 \, \text{m/s}^2 \), so \( a_c = 1.75 \times 9.81 \, \text{m/s}^2 \).

First, calculate the centripetal acceleration:

\[
a_c = 1.75 \times 9.81 = 17.1675 \, \text{m/s}^2
\]

Now, rearrange the centripetal acceleration formula to solve for \( v \):

\[
v = \sqrt{a_c \times r}
\]

Substitute the values of \( a_c \) and \( r \):

\[
v = \sqrt{17.1675 \times 15.0}
\]

Let me calculate the speed.The speed of the riders is approximately \( 16.05 \, \text{m/s} \).

Would you like further details or have any other questions?

Here are five related questions to deepen your understanding:
1. What is the relationship between centripetal force and acceleration?
2. How does increasing the radius of the circular path affect the riders' speed?
3. If the acceleration was increased to \(2 \, g\), what would the new speed be?
4. How is centripetal acceleration different from tangential acceleration?
5. What would happen to the speed if the radius was halved but the same acceleration applied?

**Tip:** In circular motion, the centripetal acceleration always points toward the center of the circle and is responsible for changing the direction of the velocity, not its magnitude.

A fairground ride spins its occupants inside a large teacup. If the radius of the circular path is 15.0 m and the acceleration is 1.75 g, what is the speed of the riders?

Discover how to calculate the speed of riders on a fairground ride moving in circular motion with a radius of 15.0 m and centripetal acceleration of 1.75 g. Learn the step-by-step solution using key formulas from physics, ideal for students in Grades 10-12.

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