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Vertical Circular Motion & Centripetal Acceleration 
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#1
Feb808, 09:19 PM

P: 12

1. The problem statement, all variables and given/known data
1) A car travels at a constant speed around a circular track whose radius is 2.1 km. The car goes once around the track in 410 s. What is the magnitude of the centripetal acceleration of the car? 2. Relevant equations a=(v^2)/r 3. The attempt at a solution a=((2100/410)^2)/2100=0.01249  wrong answer  1. The problem statement, all variables and given/known data 2)A roller coaster at an amusement park has a dip that bottoms out in a vertical circle of radius r. A passenger feels the seat of the car pushing upward on her with a force equal to five times her weight as she goes through the dip. If r = 20.5 m, how fast is the roller coaster traveling at the bottom of the dip? 2. Relevant equations Fmg = m(v^2/r) 3. The attempt at a solution v^2 = (20.5)(9.8) v = 14.17 m/s  wrong answer  Please help me, I don't know what I'm doing wrong. 


#2
Feb808, 09:40 PM

P: 93

I believe you miscalculated the velocity in the first question. You probably want to give this another look.
In the second question, try rewriting F mg = m(v^2)/r all in terms of m, g, v, and r. Then go through and see what you can cancel. 


#3
Feb808, 09:56 PM

P: 12

Thank you so much for replying.
For the 1st problem: 2.1 km = 2,100 m velocity = 2,100 m / 410 s = 5.12 m/s (did anything went wrong in this step?) a = (v^2)/r a = (5.12^2)/2100 = 0.01248  For the 2nd problem: F  mg = m(v^2)/r Force is 5 times the weight, so F=5m 5m  mg = m(v^2)/r m(5g) = m(v^2)/r rm(5g)/m = m(v^2)/m r(5g) = v^2 (20.5)(59.8) = v^2 98.4=v^2 (Do I ignore the negative sign?) 


#4
Feb808, 10:21 PM

P: 93

Vertical Circular Motion & Centripetal Acceleration
For the first question:
2.1 km is the radius of the track. You need to use this information to find the entire length of the track (more specifically, the circumference). Then use this information to find the velocity. For the second question: As you said, the force is 5 times the weight. However, this is not F = 5m. It should be F = 5mg because W = mg (the MASS is m, but the WEIGHT is mg). I hope this makes sense; if not, I'll be up for a while and would be happy to try to help further. 


#5
Feb808, 10:34 PM

P: 12

Thank you for your help.
I got the 2nd problem. The answer came out to be 28.35 m/s. For the 1st problem: a = (v^2)/r r = 2100/(2*3.14) = 334.39 velocity = 334.39m/410s = 0.8156 m/s? then the acceleration would be a= 0.8156/410 = 0.001989 m/s^2? 


#6
Feb808, 10:39 PM

P: 93

Congratulations on getting the answer to the second question.
Now, let's try to iron the wrinkles in the first problem. The problem gives you the value for r. This is 2.1 km = 2,100 m. It also gives you the time it takes for the car to make one full lap around the track. To get the velocity, you need to find the distance the car travels in the time it took to complete one lap. Thus, you need to find the length of the track and then divide this number by the time it took to traverse the track. How do you think you will find the length of the track? HINT: The circumference of a circle = 2*pi*r 


#7
Feb808, 10:47 PM

P: 12

So, the circumference is (2)(pi)(2100) = 13,194.69 m
the car traveled for 410 s, 13194.69/410 = 32.18 m/s? EDIT: I got it! Thank you so much. You are a lifesaver! 


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