Question

A car in an amusement park ride rolls without friction around a track (Fig. P7.42). The hB car starts from rest at point A at a CR height h above the bottom of the loop. Treat the car as a particle. (a) What is the minimum value of h (in terms of R) such that the car moves around the loop without falling off at the top (point B)

Answer 1

The minimum value of h (in terms of R) such that the car moves around the loop without falling off at the top is h/R >= 1/2.

Step-by-step explanation:

To determine the minimum value of h (in terms of R) such that the car moves around the loop without falling off at the top (point B), we need to consider the conservation of energy. At point A, the car has potential energy equal to mgh, where m is the mass of the car and g is the acceleration due to gravity. At point B, the car has both potential energy and kinetic energy. The minimum value of h occurs when all the potential energy at point A is converted into kinetic energy at point B.

Using the conservation of energy, we can equate the potential energy at A to the sum of potential energy (zero) and kinetic energy at B:

mgh = mghB + (1/2)mvB^2

Simplifying the equation:

gh = ghB + (1/2)vB^2

Since the car is not experiencing any friction, the net acceleration at the top of the loop is given by a = g + (vB^2/R).

As the car moves without falling off at the top of the loop, the net acceleration at the top must be greater than or equal to zero:

g + (vB^2/R) >= 0

Substituting g + (vB^2/R) for a in the previous equation and solving for h:

(ghB - gh)/(vB^2/R) >= 1/2

Since the car is at minimum h, hb = 0, and the equation becomes:

gh/(vB^2/R) >= 1/2

Simplifying, we get:

h/R >= 1/2

Therefore, the minimum value of h (in terms of R) such that the car moves around the loop without falling off at the top (point B) is h/R >= 1/2.

Answer 2

`h>(5)/(2)R`

Step-by-step explanation:

Given that

Height = h

Radius = R

From energy conservation

`KE_A+U_A=KE_B+U_B`

At point B

The minimum speed to complete the the circle

`V_B=√(gR)\ m/s`

So the kinetic energy at point B

`KE_B=(1)/(2)mV^2`

`KE_B=(1)/(2)mgR`

`KE_A+U_A=KE_B+U_B`

`0+mgh=(1)/(2)mgR+2mgR`

Without falling off at the top (point B)

`0+mgh>(1)/(2)mgR+2mgR`

`mg(h-2R)>(1)/(2)mgR`

`g(h-2R)>(1)/(2)gR`

`h>(5)/(2)R`