# Phys Unit 4 - Swings

1. Jan 29, 2012

### jsmith613

1. The problem statement, all variables and given/known data
Question 6:
http://www.xtremepapers.com/Edexcel/Advanced%20Level/Physics/2008%20Jan/6754_01_que_20080121.pdf [Broken]

"On one type of theme park ride, a...."

2. Relevant equations

3. The attempt at a solution
So at the bottom of the loop the ride is at its fastest
GPE = KE
mgh = 0.5*m*v2
max speed = 18.8

The net force is mv2/r = Tension - mg

How do I continue???

Last edited by a moderator: May 5, 2017
2. Jan 29, 2012

### Staff: Mentor

Consider the forces on the rider. (Not tension, but what?)

3. Jan 29, 2012

### jsmith613

reaction force from seat?

4. Jan 29, 2012

### Staff: Mentor

Right. The reaction or normal force from the seat.

5. Jan 29, 2012

### jsmith613

ok I think I am working on it...

6. Jan 29, 2012

### jsmith613

so
R = mv^2/r + mg
R/mg = g-force
g-force = 3
right?

7. Jan 29, 2012

### jsmith613

but the next question 6b) is

Two students in the queue are having a discussion.
A says: “If they made a new ride twice as big the g-force at the bottom would be
amazing!”
B says: “I think the g-force wouldn’t be any different.”
With reference to your calculation, explain which student is correct.

If we look at our equation r (radius) is clearly a factor
but the answer says B is correct>
how?

8. Jan 29, 2012

### jsmith613

or would this cancel out due to a higher max speed?

9. Jan 29, 2012

### Staff: Mentor

Right.

10. Jan 29, 2012

### Staff: Mentor

mv2 also depends on the radius. Express that term in terms of the radius and see what happens.

11. Jan 29, 2012

### jsmith613

yes they then cancel out
so (2g+g)/g = g-force
3 = g-force

12. Jan 29, 2012

### Staff: Mentor

Good.

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