In a loop-the-loop ride a car goes around a vertical, circular loop at a constant speed. The car has a mass m = 268 kg and moves
with speed v = 16.24 m/s. The loop-the-loop has a radius of R = 10.5 m. 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?
Before we go through the questions, we need to calculate and determine some values first.
r = 11.5 m <span>m = 280 kg </span> <span>Centripetal force = m x v^2/r = 280 x (17.1^2/11.5) = 7119.55 N </span> 1) What is the magnitude of the normal force on the care when it is at the bottom of the circle.
<span>Centripetal force + mg = 7119.55 + (280 x 9.8) = 9863.55 N </span>
<span>2) What is the magnitude of the normal force on the car when it is at the side of the circle. </span>
<span>Centripetal force = 7119.55 N </span>
<span>3) What is the magnitude of the normal force on the car when it is at the top of the circle. </span>
<span>Centripetal force - mg = 7119.55 - (280 x 9.8) = 4375.55 N </span>
<span>4) What is the minimum speed of the car so that it stays in contact with the track at the top of the loop. </span>
Gravity is one major force that creates tides. In 1687, Sir Isaac Newton explained that ocean tides result from the gravitational attraction of the sun and moon on the oceans of the earth (Sumich, J.L., 1996).
