0 A loop-the-loop arrangement has a ramp of height h and a circular track of rad

Question

0 A loop-the-loop arrangement has a ramp of height h and a circular track of radius r. A car with mass m starts at the top and is released from rest. Assume that there is no friction. Let h 32.0 m, 10.00 m, and m-120 kg A) Find the speed of the car (in m/s) at the bottom of the circular track (location 1) a) 23.4 b) 28.0 c) 26.6 25.0 B) Find the speed of the car (in m/s) at the right-most point of the circle (location 2) 2n V20.8 ,b) 18.3 ,c) 25.0 ,d) 23.0 . C) Find the speed of the car (in m/s) just before it leaves the circle (at location 3) b) 15.3 ) ,c) 10.8 ,d) 21.7-2nN's vho10ww Find the total acceleration of the car in m/s?) at location 2. (Don't forget about the 31.8 18.8 D) tangential component) 10 a) 79.0b) 59.6c) 44.2 a52-219 : 43.2 20. E) Find the normal force (in N) acting on the car at location 3. a) 125 X3920 c) 2940 ) 1646 2 2 2932-10) 09(31-20)

Explanation / Answer

Given,

h = 32 m ; r = 10 m ; m = 120 kg ;

A)from conservation of energy

1/2 m v^2 = m g h

v = sqrt (2 gh) = sqrt (2 x 9.81 x 32) = 25 m/s

Hence, (c)25 m/s

B)Now we have

h' = 32 - 10 = 22

v = sqrt (2 g h) = sqrt (2 x 9.81 x 22) = 20.8 m/s

Hence, (a)20.8 m/s

C)At the top

h' = 32 - 20 = 12

v = sqrt (2 x 9.81 x 12) = 15.3 m/s

Hence, (b)15.3 m/s

D)a = v^2/R

a = 20.8^2/10 = 43 m/s^2

g = 9.81

a2 = sqrt (43 + 9.8^2) = 44.2 m/s^2

Hence,(c)44.2 m/s^2

Related Questions

Navigate

• Browse (All)
• Browse
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.