# Homework Help: Centripetal Acceleration of roller-caster vechicle

1. Apr 1, 2006

### dontcare

A roller-caster vechicle has a mass of 500 kg when fully loaded with passengers (Fig P7.28) (a) If the vechile has a speed of 20.0 m/s at point A, what is the force of the track on the vehicle at this point? (b) What is the maximum speed the vehicle can have at point B in order for gravity to hold it on the track?

a) m = 500 kg
$$V_{t} = 20 m/s$$
$$F_{t} = ?$$
r = 10 m

$$a_{c} = \frac{v^2}{r} = 40 m/s^2$$
$$F = m( g + a_{c} ) = 2.49 * 10^4 N$$

b) m = 500 kg
$$V_{t} = ?$$
$$F_{t} = ?$$
r = 15 m

$$F_{T} = n - F_{c} = m(g - a_{c} )$$
$$0 = m(g - a_{c} )$$
$$g = a_{c}$$
$$g = \frac{V_{t^2} }{r}$$
$$V_{t} = \sqrt{9.8 m/s^2 * 15 m} = 12.1 m/s$$

Could someone explain part (b). Why is there no acting F on the vehicle when it has reaced the maximum point on the roller coaster. Take a look at (Fig. P7.28).

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2. Apr 1, 2006

### nrqed

It's really a bad habit to use a centripetal force in the equations... It leads to a LOT of confusion!! (I am not sure if you learned that from your prof or a book ??). There *IS* a force at the top!!

The real equation is

$\sum F_y = n- mg = m a_y$. The only forces are the normal force and gravity. And $a_y = - v^2/r = - a_c$.

The only thing that is special when the object reaches the maximum speed is that the *normal force* is zero, leaving only gravity. So the equation reduces to

$-mg = - a_c$.

This gives the same answer but the physics is much more clear! There IS a force and it`s gravity. What is special at the top is that the normal force is zero. No need to introduce a centripetal force (the centripetal force is really NOT a new force, but the sum of all the forces acting toward the center of the circle!!)

Patrick

3. Apr 1, 2006

### dontcare

Thank you. My prof. told us to solve the question this way. He did mention that the equation he gave us for centripetal force was incorrect but he didn't tell the class why.