A roller coaster is created with a hill of height 125 m and a loop with radius 4

Question

A roller coaster is created with a hill of height 125 m and a loop with radius 45m. To make the crowds swoon, a new cart is brought in: a hollow sphere of mass 250kg and radius 2m, capable of carrying four passenger of total mass, 400kg.

The hollow 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 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 it change or is it still the same?)

1.5 If the hollow sphere is replaced by a hollow disk with same mass and radius as the sphere, does the

speed in 1.4 change?

1.6 What is the maximum height of a loop for a rolling, hollow sphere? Why is it different than that of a sliding object?

Explanation / Answer

Given: height of the hill (h) = 125 m ; radius of the loop (r) = 45 m ; mass of the hollow sphere (m) = 250 Kg

and total mass of people (M) = 400 Kg

1.1 Initial energy of the system (cart+people) E1 = (m+M)*g*h = 650*10*125 = 812500 J

1.2 Let the speed at the bottom be v then applying energy conservation between top and bottom of the hill

E1 = 1/2*(m+M)*v2   or v = [2*E1/(m+M)]0.5  = (2*812500/650)0.5 = 50 m/s

1.3 The net difference between height of the top of the hill and the height at the top of the roller coaster (h1) = h - 2r or h1 = 125 - 2*45 = 35 m

Hence applying energy conservation we have: decrease in potential energy = increase in kinetic energy

or (m+M)*g*h1 = 1/2*(m+M)*v12   where v1 is the velocity at the top of roller coaster

hence v1 = [2*g*h1]0.5 = (2*10*35)0.5 = 26.45 m/s

1.4    The minimum speed to get around the loop depends upon the speed at the top of roller coaster where the weight must be accounted for the centripetal force i,e (m+M)*g = (m+M)*vmin2/r where vmin is the minimum velocity to get around the loop.

Hence vmin = (rg)0.5   so

1.5     No it doesn't change by replacing the cart as it only depends on the radius of the loop.

1.6 For a rolling hollow sphere the total kinetic energy (KE) is given by:

KE = KE(translation) + KE(rotation) = 1/2*(m+M)*v2 + 1/2*I*2   where I is the moment of inertia of the cart+people system = 2/5*m*r2 + M*r2   and is the angular velocity = v/r (for pure rotation)

so KE = 7/10*m*v2 + M*v2

Now for the maximum height of the loop we need to consider the minimum velocity case at the top of the loop:

Applying energy conservation between the loop's top and the top of the hill:

E1 = KE(vmin) + (m+M)*g*hmax   or 812500 = 0.7*250*45*10 + 400*45*10 + 650*10*hmax

or hmax   = 553750/6500 = 85.2 m

The hmax will be different for sliding object as then there would be contribution coming from the rotation of the object.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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