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Chapter 3: Problem 49

The singapore Flyer is the world's largest Ferris wheel. Its diameter is \(150 \mathrm{m}\) and it rotates once every 30 min. Find the magnitudes of (a) the average velocity and (b) the average acceleration at the wheel's rim, over a 5.0 -min interval.

Short Answer

Expert verified
The average velocity of the point on rim of the Ferris wheel for a 5 minute interval is 15.708 m/min and the average acceleration of the point on rim of the Ferris wheel for a 5 minute interval is \( 3.292 \times 10^{-3} m/min^2 \).

Step by step solution

01

Find the circumference of the wheel

We know the diameter of the ferris wheel is 150 m. It's like a circle, and a circle's circumference can be calculated by the formula \(C = \pi d\). Substituting \(d = 150m\) gives us \(C = \pi \times 150m\). That means the circumference of the wheel, the distance a point at the rim travels in one full rotation, is \(C = 471.24m\)
02

Calculate the average velocity

The average velocity is calculated as the total distance travelled divided by the time it took to travel that distance. We know a point on the rim travels a distance equal to circumference of the wheel \(C = 471.24m\) for every 30 minutes rotation. However, we're asked to find the average velocity for a 5-minute interval. So, first, we need to adjust the total distance travelled in a 5-minute duration, giving us \(d = C \times \frac{5min}{30min} = 78.54m\). Then, average velocity \(v\) for 5 minutes interval is calculated as \(v = \frac{d}{5min} = 15.708 m/min\).
03

Calculate the average acceleration

The only acceleration occurring here is centripetal, or radial, acceleration due to the change in direction of the velocity vector which occurs because the rim is moving in a circle. The average acceleration is calculated as the change in velocity divided by the time it took to change that velocity. But as we know, in centripetal motion, only the direction of the velocity changes not the magnitude. Hence, the centripetal acceleration is \( a = \frac{v^2}{r} \) where \(v\) is velocity and \(r\) is radius of the circle (half of the wheel's diameter, or 75m). Substituting the values, gives us \( a = \frac{(15.708 m/min)^2}{75m} = 3.292 \times 10^{-3} m/min^2 \).

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