EXAMPLE 3.0 A high-speed carnival ride In this example the motion is in a vertic

Question

EXAMPLE 3.0 A high-speed carnival ride In this example the motion is in a vertical circle, Passengers in a carnival ride travel in circle ride moves at with radius m (Figure 3.24) The time T 4.0 a constant speed and makes one complete circ in a s, What is the acceleration of the passengers? ON SET UP Figure 3.25 shows our diagram. T 4.0 s R 50n A FIGURE 3.25 Our diagram for this problem. FIGURE324 Circular motion in a carnival ride. SOLVE We again use Equation 3.12 a v IR. To find the speed v, we use the fact that a passenger travels a distance equal to the circum- ference of the circle (2mR) in the time Tfor one revolution: 2 m) 2 mR 4.0 s 79 m/s. The centripetal acceleration is 12 m/s and "R" 5.0m REFLECT As in Example 38, the direction of a is toward the center of the circle. The magnitude of ais greater than g the accel due to leration gravity, so this is not a ride for the faint-hearted. (But some roller coast. ers subject their passengers to accelerations as great as 4g) Practice Problem: If the ride increases in speed so that T 20s. what is and? This question answered by using proportional rea- can be soning, without much arithmetic. Answer: 49 m/s

Explanation / Answer

v = 2*pi*R / T

v = 2*3.14*5 / 2

v = 15.708 m/s

a (rad) = v^2 / R

a (rad) = 15.708^2 / 5

a (rad) = 49.3 m/s^2 = 49 m/s^2

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