- #1

- 322

- 45

- Homework Statement
- Show that on a roller coaster with a circular vertical loop

(Fig. 43), the difference in your apparent weight at the top of

the loop and the bottom of the loop is 6 g’s—that is, six times

your weight. Ignore friction. Show also that as long as your speed is above the minimum needed, this answer doesn’t depend on the size of the loop or how fast you go through it.

- Relevant Equations
- curvilinear motion + conservation of energy + newton's laws

Hello there, I have tried the problem but don't get a different of 6g's as I am supposed to. I am not sure whether I interpreted the problem in the correct way, but I would love some feedback/hints on what went wrong in my solution, thanks in advance.

Solution:

SITUATION DRAWINGS + FBDS

so based on the situations described in the situational picture:

For ##F_n## at ##2##:

##F_n - mg = ma_n## where ##a_n = \frac{v_2^2}{R}##

--> ##F_n = m(\frac{v_2^2}{R} + g)## ##(1)##

For ##F_n## at ##3##:

##F_n + mg = m*\frac{v_3^2}{R}##

##F_n = m(\frac{v_3^2}{R} - g)## ##(2)##

Then moving on to finding those velocities we use conservation of Mechanical energy for both of the situations (2 and 3):

##E_1 = E_2##

##\frac{mv_1^2}{2} + mgy_1 = \frac{mv_2^2}{2} + mgy_2##

##0 + mgh = \frac{mv_2^2}{2} + 0##

##v_2 = \sqrt{2gh}##

for ##v_3##:

##E_1 = E_3##

##mgh + 0 = mg(2R) + \frac{mv_3^2}{2}##

## v_2 =\sqrt{2gh - 4gR}##

substituting these in ##(1)## and ##(2)##:

##F_{n,2} = m(\frac{2gh}{R} + g) = \frac mR (gR + 2gh)##

##F_{n,3} = m(\frac{2gh - 4gR}{R} - g) = \frac mR (2gh - 4Rg - Rg) = \frac mR (2gh -5Rg) ##

##\Delta F_n = F_{n,3} - F_{n,2} =\frac mR (2gh -5gR) - \frac mR (gR + 2gh)##

## = \frac mR (gh - 6gR)##

This is my answer which is definitely not 6g's (note, this is only part one of the problem) i.e. not

Solution:

SITUATION DRAWINGS + FBDS

so based on the situations described in the situational picture:

For ##F_n## at ##2##:

##F_n - mg = ma_n## where ##a_n = \frac{v_2^2}{R}##

--> ##F_n = m(\frac{v_2^2}{R} + g)## ##(1)##

For ##F_n## at ##3##:

##F_n + mg = m*\frac{v_3^2}{R}##

##F_n = m(\frac{v_3^2}{R} - g)## ##(2)##

Then moving on to finding those velocities we use conservation of Mechanical energy for both of the situations (2 and 3):

**Note:**I assumed ##v_1 = 0##, I don't know if that is permitted as it doesn't say it and perhaps the mistake lies in this one?##E_1 = E_2##

##\frac{mv_1^2}{2} + mgy_1 = \frac{mv_2^2}{2} + mgy_2##

##0 + mgh = \frac{mv_2^2}{2} + 0##

##v_2 = \sqrt{2gh}##

for ##v_3##:

##E_1 = E_3##

##mgh + 0 = mg(2R) + \frac{mv_3^2}{2}##

## v_2 =\sqrt{2gh - 4gR}##

substituting these in ##(1)## and ##(2)##:

##F_{n,2} = m(\frac{2gh}{R} + g) = \frac mR (gR + 2gh)##

##F_{n,3} = m(\frac{2gh - 4gR}{R} - g) = \frac mR (2gh - 4Rg - Rg) = \frac mR (2gh -5Rg) ##

##\Delta F_n = F_{n,3} - F_{n,2} =\frac mR (2gh -5gR) - \frac mR (gR + 2gh)##

## = \frac mR (gh - 6gR)##

This is my answer which is definitely not 6g's (note, this is only part one of the problem) i.e. not

*'Show also that as long as your speed is above the minimum needed, this answer doesn’t depend on the size of the loop or how fast you go through it.'*