Loop-the-Loop In a loop-the-loop ride a car goes around a vertical, circular loo
ID: 3281274 • Letter: L
Question
Loop-the-Loop In a loop-the-loop ride a car goes around a vertical, circular loop at a constant speed. The car has a mass m . 205 kg and moves with speed v . 15.64 m/s. The loop-the loop has a radius of R-9.5 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.) What is the magnitude of the normal force on the car when it is at the side of the circle (moving vertically upward)? What is the magnitude of the normal force on the car when it is at the top of the circle? 4Compare the magnitude of the cars acceleration at each of the above locations: Oabottom-aside" atop Oabottom aside atop abotto aside> atogp Submitted: Thuriday, October 26 at 300 PM Feedback: Correct 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
Given that
Mass m = 205 kg
Speed v =15.64 m/s
R = 9.5 m
we know that
Centripetal force = m x v^2/r = 205 x 15.64^2/9.5 =5278.42 N
1) Centripetal force + mg =5278.42 + (205 x 9.8) =7287.42 N
2) Centripetal force = 5278.42 N
3) Centripetal force - mg = 5278.42 -(205 x 9.8) =3269.42 N
4) a = v^2/r so always same.
a bottom = a top = aside
5) sq-root(gr) =sq-root(9.8 x 9.5) = 9.65 m/s
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.