In a loop-the-loop ride a car goes around a vertical, circular loop at a constan

Question

In a loop-the-loop ride a car goes around a vertical, circular loop at a constant speed. The car has a mass m = 218 kg and moves with speed v = 15.73 m/s. The loop-the-loop has a radius of R = 10.5 m.

1)What is the magnitude of the normal force on the care when it is at the bottom of the circle? (But as the car is accelerating upward.)

2)What is the magnitude of the normal force on the car when it is at the side of the circle (moving vertically upward)?

3)What is the magnitude of the normal force on the car when it is at the top of the circle?

4)Compare the magnitude of the cars acceleration at each of the above locations:

a)abottom = aside = atop

b)abottom < aside < atop

c)abottom > aside > atop

5)What is the minimum speed of the car so that it stays in contact with the track at the top of the loop?

Explanation / Answer

1)At bottom the free body diagram equation will be

N-mg=(mv^2)/r; v=velocity at bottom

2)At side it will be

N=mv^2/r; v= velocity at sides

3) At top Mg+N=mv^2/r;

v=velocity at top

Now u can use the Energy conservation to solve the probem woth thw given values.

Any how its given constant speed so every put v=15.73

4)

Now to get the acceleration here constant speed indicates uniform circular motion so

magnitude of accelereration i.e normal acceleration is equal at all points.

5)Min speed is square root of (gr) at the top and at the bottom a value of sqare root of (5gr) is neeeded to complete full circle.

Put the values of g and R accordinglky.

All the best for ur soln.

Feel free to ask doubts.

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