Need some expertice on rotational motion

[physics noob]

Q) a 250kg roller coaster travels on along track as shown*** at point A and B the coaster is traveling along circular paths. the speed at point A is 20m/s what is the max speed the coaster can have at B and still remain on the track.
*** ill do my best to describe this track...basically looks like one half circle the bottom half, meaning concave up, with point A at the bottom, Radius = 10m...then that half circle is attached to another. bigger half circle but this time the upper half, meaning concave down, with point B on top. Radius equals 15m...imagine drawing an S on a paper then turning it 90 degrees to the left, that's what it looks like... ok so i had to find the force of n at point A,,, i got that part, but now, it wants to know max speed at point B, so i thinks to myself this...
(m)(Ac)= normal force + mg at point B

so 250 * ((V)^2)/15 = normal force + 250(9.8)

how do i calculate n? i did try working backwards...using the velocity at point A and trying to find speed where the 2 half circles meet, then trying to find velocity at point B...not sure if that's how to do this problem, but if it is, then i also need help finding the velocity for point B... any help is greatly appriciated

 

[physics noob]

FALSE ALARM... i just busted the MEi +Wnc= MEf equation on this mofo and got my velocity at point B, man this was a tricky problem...sorry to spam the boards......IM THE SMARTEST MAN ALIVE

 

[thepatientmental]

To calculate the maximum speed at point B, you will need to use the concept of centripetal force. This is the force that keeps an object moving in a circular path. In this case, the roller coaster is traveling along a circular path at both point A and B.

To calculate the maximum speed, you will need to equate the centripetal force at point B to the maximum force that the roller coaster can withstand without flying off the track. This maximum force is equal to the weight of the roller coaster (mg) plus the normal force at point B.

The formula for centripetal force is Fc = mv^2/r, where m is the mass of the roller coaster, v is the velocity, and r is the radius of the circular path.

At point B, the radius of the circular path is 15m. The mass of the roller coaster is 250kg, and the force of gravity is 250(9.8) = 2450N.

So, the equation for the maximum force at point B is:

Fc = mv^2/r = 250v^2/15 = 2450 + n

Where n is the normal force at point B. To solve for v, you will need to rearrange the equation:

250v^2/15 = 2450 + n

v^2 = (2450 + n) * 15/250

v = √ [(2450 + n) * 15/250]

Now, to find the value of n, you can use the equation you already calculated for the normal force at point A:

n = (m)(Ac) - mg = 250 * 20^2/10 - 250(9.8) = 9800N

So, the equation for the maximum speed at point B becomes:

v = √ [(2450 + 9800) * 15/250] = √ 686 = 26.2 m/s

Therefore, the maximum speed the roller coaster can have at point B and still remain on the track is 26.2 m/s.

I hope this helps clarify the problem for you. Remember to always use the concept of centripetal force when dealing with rotational motion problems involving circular paths. Good luck!