Question

A roller coaster at a popular amusement park has a portion of the track that loops. Assuming that the roller coaster is frictionless, find the velocity at the top of the loop.

V1 = 0 m/s
height of the starting point (where V1 = 0) = 70m
height of the loop = 30m
V2 = ?

Answer 1

Condition: A roller coaster at a popular amusement park has a portion of the track that loops. Assuming that the roller coaster is frictionless, find the velocity at the top of the loop.

$V_1=0 (\frac{m}{s})$

height of the starting point (where V1 = 0) $h_1=70 (m)$  

height of the loop  $h_2=30(m)$

$V_2-?$

Solution: According to the law of conservation of energy:

$K_1+P_1= K_2+P_2$, (1)

where $K_1$ is a kinetic energy in the starting point, $K_1=0$, because $V_1=0$; $P_1=mgh_1$ is a potential energy in the starting point; $K_2=\frac{mV^2}{2}$ is a kinetic energy of the loop; $P_2=mgh_2$ is a potential energy of the loop.

Rewrite (1) taking into account $K_1, K_2, P_1, P_2.$

$mgh_1=\frac{mV^2}{2}+mgh_2$, (2)

where m will decrease. So:

$gh_1=\frac {V_2^2}{2}+gh_2$ (3)

Now, let’s find $V_2.$

$\frac {V_2^2}{2}=gh_1-gh_2$

$\frac {V_2^2}{2}=g(h_1-h_2)$

$V_2^2=2g(h_1-h_2)$.

So

$V_2=\sqrt{2g(h_1-h_2)}$ (4)

Substitute numbers from the condition and calculate $V_2$:

$V_2=\sqrt{2*9.8(70-30)}=\sqrt{19.6*40}=\sqrt{784}=28 (\frac{m}{s})$.

Answer: $V_2= 28 (\frac{m}{s}).$