The carnival ride shown in Figure 7-18 has a 2.0 m radius and rotates once each .90 s
a. Find the speed of the rider.
b. Find the centripetal acceleration of the rider.
c. What produces this acceleration?
SOLUTIONS PROVIDED:
a. 141
b. 98
c. --Explain--
Completely stuck and I don't get the answer in any way.
a ) it is 2 *pi *2/0.9 = 13.9 m/s
b) 13.9ˆ2 /2 = 96,6 m/sˆ2 Correct?
The acceleration is proviced by the rotating wheel, and more importantly, whatever is holding the rider in place. That force is accelerating the rider inward.
a. v=2PIr/period=6.28*2/.9 how exactly did you get 141?
b. centacceleratio= v^2/r
An answer key provided, although I was in a bit of a rush so I could have possibly copied the answer down wrong...
she could have meant 14.1... just saying
Well, let's see if I can lend you a helping hand with a touch of humor!
a. Find the speed of the rider.
To find the speed of the rider, we need to determine how fast they're going around the carnival ride. Since the ride completes one rotation in 0.9 seconds, it means that in one second, they complete 1/0.9 = 1.11 rotations.
But, how far is that in terms of distance traveled? Well, the distance around a circular path is given by the formula 2πr, where r is the radius. So, in this case, the distance traveled in one second is 2π(2.0) = 12.57 meters.
Therefore, the speed of the rider is 12.57 meters per second.
b. Find the centripetal acceleration of the rider.
Now, let's move on to the centripetal acceleration of the rider. The formula for centripetal acceleration is a = v^2/r, where v is the velocity and r is the radius.
Using the speed we just found (12.57 m/s) and the radius (2.0 m), we can plug the numbers into the formula:
a = (12.57)^2/2.0 = 98.51 m/s^2
So, the centripetal acceleration of the rider is approximately 98.51 m/s^2.
c. What produces this acceleration?
Well, the name gives it away - it's the "centripetal" acceleration, which means it's directed towards the center of the circular path. In this case, the acceleration is produced by the combination of the rider's inertia (wanting to move in a straight line) and the restraining force exerted by the carnival ride. So, it's like a partnership between the rider's desires and the ride's ability to keep them on track.
No problem! I'll explain step by step how to find the answers to these questions.
a. To find the speed of the rider, we can use the formula for linear velocity (v) in circular motion. The linear velocity is the distance traveled per unit of time. In this case, the rider completes one revolution every 0.90 seconds.
The formula for linear velocity is:
v = 2πr / T
where v is the linear velocity, r is the radius of the circular path, and T is the period (time taken to complete one revolution).
In this case, the radius (r) is given as 2.0 meters and the period (T) is given as 0.90 seconds. Substituting these values into the formula, we get:
v = 2π(2.0) / 0.90
Evaluating the expression, we find:
v ≈ 4.42 m/s
Therefore, the speed of the rider is approximately 4.42 m/s.
b. To find the centripetal acceleration of the rider, we can use the formula for centripetal acceleration (a_c) in circular motion. The centripetal acceleration is the acceleration towards the center of the circular path.
The formula for centripetal acceleration is:
a_c = (v^2) / r
where a_c is the centripetal acceleration, v is the linear velocity, and r is the radius of the circular path.
In this case, the linear velocity (v) is 4.42 m/s, and the radius (r) is 2.0 meters. Substituting these values into the formula, we get:
a_c = (4.42^2) / 2.0
Evaluating the expression, we find:
a_c ≈ 48.76 m/s^2
Therefore, the centripetal acceleration of the rider is approximately 48.76 m/s^2.
c. The centripetal acceleration is produced by the net force acting towards the center of the circular path. In this case, that force is provided by the tension in the ride's restraining mechanism or any other force that acts inward and keeps the rider moving in a circular path.
I hope this helps! Let me know if you have any further questions.