Solving for Maximum Speed on a Theme Park Ride: Phys Unit 4 - Swings Question 6

[jsmith613]

Homework Statement

Question 6:
http://www.xtremepapers.com/Edexcel/Advanced%20Level/Physics/2008%20Jan/6754_01_que_20080121.pdf

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

Homework Equations

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?

 

[Doc Al]

The net force is mv²/r = Tension - mg

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

 

[jsmith613]

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

reaction force from seat?

 

[Doc Al]

reaction force from seat?

Right. The reaction or normal force from the seat.

 

[jsmith613]

ok I think I am working on it...

 

[jsmith613]

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

 

[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?

 

[jsmith613]

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

 

[Doc Al]

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

Right.

 

[Doc Al]

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

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

 

[jsmith613]

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

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

 

[Doc Al]

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

Good.