Physics HWK. Problem-Vertical Circular Motion

[shawonna23]

A roller coaster at an amusement park has a dip that bottoms out in a vertical circle of radius r. A passenger feels the seat of the car pushing upward on her with a force equal to three times her weight as she goes through the dip. If r = 16.0 m, how fast is the roller coaster traveling at the bottom of the dip?

I used the equation: v= square root of 3*r*g, but the answer was wrong. I don't know what other equation to use.

 

[ehild]

If three times her weight is what she feels and she is in a circular type system then you can find the velocity using centripetal force equations:

F = m*v^2/r

3*m*g = m * v^2/r

sqrt(3*g*r) = v

The passenger feels the normal force, which is 3mg.

(Any time when we sit, stand, ... on something we feel the normal force. On a horizontal surface, in rest, this normal force happens to be equal to mg. Not when we are in circular motion.)

There are two forces, acting on the passenger, the normal force upward and mg downward. The resultant is 2mg, and that is equal to the centripetal force. So v=sqrt(2mg)

 

[wais]

Hi there,

There are actually a few different equations that can be used to solve this problem. The equation you used, v= √(3rg), is correct, but it is important to make sure that all of the units are consistent. In this case, r is given in meters, g is the acceleration due to gravity (9.8 m/s^2), and v is the velocity in meters per second. So, plugging in the values we have:

v= √(3*16.0 m *9.8 m/s^2) = √470.4 m^2/s^2 = 21.7 m/s

Another equation that can be used for this problem is the conservation of energy formula, where the initial potential energy is equal to the final kinetic energy:

mgh = ½ mv^2

Where m is the mass of the roller coaster, h is the height of the dip (which can be calculated using the radius and the height of the dip), and v is the velocity. Solving for v, we get:

v= √(2gh)

Plugging in the values we have:

v= √(2*9.8 m/s^2 * 16.0 m) = √313.6 m^2/s^2 = 17.7 m/s

As you can see, this answer is slightly different from the one obtained using the first equation. This is because the first equation assumes that all of the potential energy is converted into kinetic energy, while the second equation takes into account any potential energy lost due to friction or air resistance.

I hope this helps! Let me know if you have any further questions.