Roller coaster at bottom of dip weight increases 50%?

AI Thread Summary
A roller coaster experiences a 50% increase in weight at the bottom of a dip with a radius of 30 meters. To find the speed at the bottom, the relationship between gravitational force and normal force must be understood, where the total downward force equals 1.5 times the gravitational force. The normal force must be greater than the weight to provide the necessary centripetal force directed towards the center of the curve. The forces acting on the car include weight and the increased normal force, which together determine the speed. Understanding these dynamics is crucial for solving the problem accurately.
BC2210
Messages
17
Reaction score
0

Homework Statement



A roller coasters weight increases 50% through the center of a dip in the track. The radius of the dip is 30m. What is the speed at the bottom?

Homework Equations



This is a uniform circular motion problem correct? But how do I calculate if I don't know the weight?

v = \omegar
Fnet = mv2/r

The Attempt at a Solution



So is weight twice that of the normal force? Or the normal force has to balance it out by increasing by 50% as well?

Or should I be thinking in terms of \omega instead of forces?
 
Physics news on Phys.org
You are on the right track ;-)
At the bottom of the curve their weight due to gravity is 'mg'
You are simply looking for the centrifugal force that equals half this.

Then the total force downward is 1.5mg
 
So right now I have F = N - (1.5mg) ...

which is the only forces acting on the car...

but I thought the force has to point to the middle of the circle?
 
N - W = (mv2)/r

but in the book it says the normal force is greater than the weight, resulting in a force pointing towards the center...
 
The normal force acting on the car must equal the force of the car on the track.
At the bottom of the curve there is weight acting down and centrifugal acting outward - which = down at the bottom of the curve.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Similar threads

Why does the speed change at the bottom of the loop-the-loop?
Replies
5
Views
2K
Speed at the top of an elliptical roller coaster loop
Replies
4
Views
3K
Minimum safe height for a roller coaster
Replies
10
Views
6K
Friction between roller coaster and track
Replies
15
Views
7K
G-force diagram on a roller coaster?
Replies
5
Views
2K
Is the Roller Coaster Speed Calculation Correct?
Replies
6
Views
3K
Roller coaster rider weight problem
Replies
3
Views
5K
Minimum Speed for a Roller-Coaster Car to Complete a Loop without Seat Belts
Replies
12
Views
3K
How Is the Force on a Child in a Roller Coaster Calculated?
Replies
3
Views
7K
What is the Velocity at the Bottom of a Roller Coaster with Given Parameters?
Replies
1
Views
1K

Hot Threads

Recent Insights

Back
Top