Calculating Roller Coaster Speed at the Bottom of a Dip

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The weight of passengers on a roller coaster increases by 50 % as the car goes through a dip with a 36 m radius of curvature.

What is the car's speed at the bottom of the dip?

Do you use the equation mv^2/r ??

how would you go about doing this problem?
 
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lmc489 said:
The weight of passengers on a roller coaster increases by 50 % as the car goes through a dip with a 36 m radius of curvature.

What is the car's speed at the bottom of the dip?

Do you use the equation mv^2/r ??

how would you go about doing this problem?
That looks like a good equation to use. What have you attempted thus far?

Just to make you aware that the PF guidelines require that you post an attempted solution, or at least detail your thoughts, before we can help you with a homework question.
 
uk=F/N
F=mg
F=mw^2r

mw^2r=N

uk=F/N

(30kg)(w^2)(2.45m)=30g/.4

somehow i don't think I am getting the right answer though... i get like 1.01 rad/s but i don't think that's right?
 
If the passengers feel as if their "weight" increases by 50% at the bottom then that means that the normal force acting up on them is going to be 50% more than the force of gravity acting down on them.

You know that the sum of the forces (in this case it would be Normal - Fg) is going to be equal to the Centripetal Force (mv^2/R).

Set up the equation and you will be able to solve for velocity.

PS- Does the problem ask for tangential of rotational velocity? I would think that for this type of problem the answer would be tangential Velocity.