Question

A roller coaster of mass 3000.0 kg starts from rest at Point A, which is 33 meters above the bottom of the coaster and proceeds to point C, which is 15 m above the bottom of the coaster. a) What is its speed at C? b) Recall that the formula for critical velocity is: If the radius of the loop is 5.0 m, will the roller coaster have enough kinetic energy at the top of the loop to satisfy the critical velocity criteria?

Answer 1

Complete answer

A roller coaster of mass 3000.0 kg starts from rest at Point A, which is 33 meters above the bottom of the coaster and proceeds to point C, which is 15 m above the bottom of the coaster. a) What is its speed at C? b) Recall that the formula for critical velocity is: $v=\sqrt{rg}$ If the radius of the loop is 5.0 m, will the roller coaster have enough kinetic energy at the top of the loop to satisfy the critical velocity criteria?

Answer:

The roller coaster has enough kinetic energy.

Explanation:

Assuming there are not dissipative forces between the cart and the rails of the roller coaster, we can use conservation of energy between A and C:

$E_A=E_B=K_A+U_A=K_C+U_C$ (1)

with K kinetic energy and U potential gravitational energy. Kinetic energy is defined as:

$K=\frac{mv^2}{2}$ (2)

with v the velocity and m the mass. If we choose zero of potential energy at the bottom of the roller coaster, the gravitational potential energy is:

$U=mgh$ (3)

with gravitational acceleration and h the height of the cart. Using (2) and (3) on (1):

$\frac{mv_A^2}{2}+mgh_A=\frac{mv_C^2}{2}+mgh_C$

Solving for VB and knowing that velocity on A is zero because the cart starts from rest:

$v_C=\sqrt{(\frac{2}{m})(mgh_A-mgh_C)}=\sqrt{2*9.81(33-15)}=18.78\frac{m}{s}$

b) Critical velocity is $v_{cr}=\sqrt{rg}$ in our case:

$v_{cr}=\sqrt{5*9.81}=7.00 \frac{m}{s}$

So, the critical kinetic energy is:

$K_cr=\frac{(3000.0)(7.0)^2}{2}=73500J$

Now we should check that the value of kinetic energy at 5m is more or equal to that value. We can use the equation used on a)

$mgh_A=K_5+mgh_5$

solving for K5:

$K_5=mg(h_A-h_5)=(3000.0)(9.81)(33-5)=824040 J$

It is bigger than Kc, so the roller coaster has enough kinetic energy.