Problem 7.46: Riding a Loop-the-loop

[annabelx4]

Homework Statement

A car in an amusement park ride rolls without friction around the track shown in the figure . It starts from rest at point at a height above the bottom of the loop. Treat the car as a particle.

What is the minimum value of (in terms of ) such that the car moves around the loop without falling off at the top (point )?

If the car starts at height 4.00 and the radius is = 25.0 , compute the speed of the passengers when the car is at point , which is at the end of a horizontal diameter.

Homework Equations

mgh = 1/2 mv^2

The Attempt at a Solution

we know that h = 2 R because it's the diameter but other than that i am lost

 

[Nate810]

. For the minimum value, I believe it is 4R. For the speed, we use the equation mgh = 1/2 mv^2. mgh = 1/2 mv^2(4)(9.8)(25) = 1/2 (20)(v^2)1960 = 10v^2196 = v^214.1 m/s = v

 

[Moetasim]

The minimum value of h, in terms of R, can be found by equating the gravitational potential energy at the top of the loop (point C) to the kinetic energy at the bottom of the loop (point A). This can be represented by the equation mgh = 1/2 mv^2, where m is the mass of the car, g is the acceleration due to gravity, h is the height at the top of the loop, and v is the speed of the car at the bottom of the loop.

Solving for h, we get h = 1/2 v^2/g. Since the car starts at height 4.00 and the radius is 25.0, we can substitute these values to get 4.00 = 1/2 v^2/g. Solving for v, we get v = √(8gh), where g = 9.8 m/s^2.

To find the speed of the passengers at point B, which is at the end of a horizontal diameter, we can use the conservation of energy principle. At point B, the car is at the same height as point A, so the potential energy is the same. However, the kinetic energy at point B is now in the form of both translational and rotational energy. We can represent this as 1/2 mv^2 + 1/2 Iω^2, where I is the moment of inertia and ω is the angular velocity.

Since the car is treated as a particle, we can simplify this to 1/2 mv^2 + 1/2 mv^2 = mv^2. Plugging in the values, we get 4.00 = 1/2 mv^2. Solving for v, we get v = √(8gh), which is the same as the speed at the bottom of the loop.

In conclusion, the minimum value of h in terms of R is 1/2 v^2/g, and the speed of the passengers at point B is √(8gh).