Consider a roller coaster as it travels near the bottom of its track as sketched in the figure below. At this point, the normal force on the roller coaster is three times its weight. If the speed of the roller coaster is 27 m/s, what is the radius of curvature of the track?

I tried the equation
r=v^2/m*gr=27^2/3*9.81, but I keep getting it wrong. Please help!

Use -3mg=mg-mv^2/r. M cancels out

so..-3g=g-v^2/r. then solve for r algebraically

To find the radius of curvature of the track, we need to use the concept of centripetal force. The normal force acting on an object moving in a curved path provides the centripetal force that keeps the object moving in that path.

In this case, the normal force on the roller coaster is three times its weight, which means that the normal force is three times the gravitational force acting on the roller coaster.

First, let's find the weight of the roller coaster. The weight is given by the equation:

Weight = mass * gravitational acceleration

Next, since the normal force is three times the weight, we can write:

Normal force = 3 * Weight

Now, let's calculate the weight of the roller coaster. We need to know the mass of the roller coaster, which is not given in the question. Assuming that we have the mass, we can substitute it into the equation to find the weight.

Once we have the weight, we can calculate the normal force using the equation above.

Finally, we can use the relationship between centripetal force, the normal force, and the radius of curvature to find the radius. The centripetal force is given by the equation:

Centripetal force = (mass * velocity^2) / radius

In this case, the centripetal force is equal to the normal force, so we can write:

Normal force = (mass * velocity^2) / radius

Rearranging the equation to solve for the radius, we get:

Radius = (mass * velocity^2) / normal force

Substituting the values we obtained for the mass, velocity, and normal force into the equation, we can find the radius of curvature.