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The radius of a loop is 12 m, what speed is necesary for the cars to stay on the track at the top of the loop?
IN an amusement park ride the radius of a loop is 12m. What speed is necessary for the cars to stay on the track at the top of the loop?
Also,
In circular motion, the acceleration is a change in direction rather than speed because
A.) it is a special type of acceleration
B.) The force is perpendicular to the instantaneous velocity
C.) it has to go in a circle
D.) gravity does not affect the motion of the object
please show work for the first problem and equations used
10 points for best answer!
thank you so much!
The "tutor" on top did not multiply the radius by 4 times... Check my work to see why it's needed
Ok, I am not sure if you are asking about the speed of the cars as they ENTER the loop on the bottom, but that's what I will use.
So, don't use centripetal force/circular motion for this. Energy is MUCH easier to use and the better. Ok. at the bottom of the loop, let's say the height is 0 right? So on the bottom, there is no potential energy, only kinetic.
To find the minimum speed, we have to state that at the top (The highest) the velocity is 0. As a result, there is no Kinetic energy on the top, only potential.
Therefore, all the kinetic energy on the bottom must be used on the top...
KE = PE
(1/2) m v^2 = m g h
1/2 v^2 = 9.8 (24) (Mass cancels out)
Solve:
velocity necessary is 21.689 m!
As for your second question: the answer is B as the acceleration's direction (Sometimes up, down, right, left) is going to change as the car will always have a centripetal force inwards.
EX: when the car is on the right hand side of the loop, there is going to be some acceleration towards the center as the Force pushes in, but the car will continue to go in its direction at a given time.....
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