A roller coaster with a hill if height 125m and a loop with radius 45m. A new ca

Question

A roller coaster with a hill if height 125m and a loop with radius 45m. A new cart is brought in: a HALLOW SPHERE of mass 250kg and radius 2m, capable to carrying 4 passengers of total mass 400kg. The hallow sphere is capable of carrying passengers around the roller coaster separately. Friction although present is mostly negligible. 1.1 what is the initial energy of the system if the cart is released from rest? 1.2 what is the speed of the cart when it reaches the bottom of the hill? 1.3 what is its speed at the top of the roller coaster loop? 1.4 what is the minimum speed to get around the loop? (Does is change it is it will the same?) 1.5 if the Hallow sphere is replaced by a HALLOW DISK with the same mass and radius as the sphere, does the speed in "1.4" change? 1.6 what is the maximum height of the loop for a rolling, hallow sphere? Why is it deferent than that of a sliding object? THANK YOU!

Explanation / Answer

1.1)

E1 = m*g*h = 250*9.8*125

E1 = 306250 J

1.2)

at the bottom

E2 = 0.5*I*w^2 + 0.5*m*v^2

I = (2/3)*m*r^2

w = v/r

E2 = 0.5*(2/3)*m*v^2 + 0.5*m*v^2

E2 = E1

0.5*(2/3)*250*v^2 + (0.5*250*v^2)= 306250

v = 38.34 m/s

+++++++

1.3)

E3 = 0.5*(2/3)*m*v^2 + 0.5*m*v^2 + m*g*2*R

E3 = E1

0.5*(2/3)*m*v^2 + 0.5*m*v^2 + m*g*2*R = m*g*h

(0.5*(2/3)*v^2) + (0.5*v^2) + (9.8*2*45) = 9.8*125

v = 20.3 m/s

1.4)

vmin = sqrt(gR) = sqrt(9.8*45) = 21 m/s

YES

1.5)

NO

1.6)

for V = 38.34 m/s

V = sqrt(gR)

38.34 = sqrt(9.8*R)

R = 150 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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