# A few simple theory questions.

1. Nov 23, 2004

### DLxX

Would someone mind answering these theory questions for me so that I can understand it.

3. Suppose a car moves at constant speed along a mountain road. At what places does it exert the greatest and least forces on the road: (a) at the top of a hill, (b) at a dip between two hills, (c) on a level stretch near the bottom of a hill?

5. A bucket of water can be whirled in a vertical circle without the water spilling out, even at the top of the circle when the bucket is upside down. Explain?

2. Nov 23, 2004

### mikezietz

3. a) least force on the road because as it is going up, it has a magnitude upwards taking away from g, the inertia makes the car want to keep going up reducing the weight on the road
b) greatest force because the magnitude will be aimed downards and because of inertia, it has more force on the road.
c) should have the same force as if just driving down a flat road.

5. the magnitude of force up overcomes mg or the gravitational force, and then some to make the water go upwards, when the bucket is down the magnitude of force on the water adds to mg or the gravitational force.

maybe an advisor etc. can give a more educated explination

3. Nov 23, 2004

### Sirus

I do not think you explained this very clearly. In 3, what has a magnitude going upwards? How does the inertia reduce the weight?

The truth is that when the road is not flat, the car undergoes something like circular motion, although it is not necessarily traveling in a circle. As the car comes to a dip, a force must exist to change the direction of motion of the car from down to flat to up (i.e. a centripetal force must exist). This force must be applied by the ground to the car since it must point upwards, so the normal force applied by the car to the ground must be greater than before the dip. Also note that when the car is traveling down an incline, the normal force is less than when it is traveling on a flat road, since part of the gravitational force on the car now translates into a force parallel to the incline. Because of this, the case of the car coming over the top of a hill is more difficult. Because the car has momentum in a direction tangential to the (theoretical) curvature of the hill, the normal force should be reduced, but now the car is no longer on an incline, so $F_{g}=F_{normal}$. This makes it hard to compare the normal force here to the normal force on the incline just before the hilltop, but the normal force at the top of the hill will still be less than the normal force on a flat surface.

5. Mike, remember that centripetal force always points towards the center of the circle. At the top of the circular path of the water in the bucket (where the bucket is upside down), the net force on the water points straight downward. What keeps it in the bucket is the fact that its instantaneous velocity points in a direction tangential to the curvature of the circular path it is following, as discussed above. At that moment, if you were to let the bucket go, it and the water would both travel in a direction tangential to the circle at the point of release.

4. Nov 24, 2004

### Staff: Mentor

Mike's explanations are correct - but allow me elaborate.

When the car is traveling up a hill, it has a upward component of velocity. It would continue to travel in that direction if there were no gravity. However when the road curves over the hill and then downward, gravity pulls the car downward. It's the same behavior in a roller coaster.

If you drive fast enough over a steep hill that then drops just as steeply, one might find one's car momentarily airborne. [Do not try that though. It's not safe]

Conversely, as the car is descending down a hill, it has a downward velocity component, and as it reaches the bottom (dip) that velocity component is changed gradually from down to zero (if the road is flat) or up if the road rises. In that case the road pushes back on the car as the car changes direction. Acceleration is the change in velocity / unit time, so at constant speed, and object is accelerating if it is changing direction. Which leads into 5.

As a bucket spins, its contents are held in place by centrifugal force (considered a fictitious force). If one lets go the bucket and contents travel in a straight line that is tangent to the circular arc.

The cord attached to the bucket is in tension and that is the application of centripetal force that keeps the bucket and contents traveling in a circle. If the bottom of bucket failed, the water would fly out in a trajectory tangent to the circular arc, and the bucket would still travel in the arc contrained by the cord.

In order for the water to stay in the bucket or for the bucket not to fall toward the holder (as a result of the force of gravity), it must be traveling with a minimum angular velocity.

http://scienceworld.wolfram.com/physics/CentrifugalForce.html

http://scienceworld.wolfram.com/physics/CentripetalForce.html

5. Nov 24, 2004

### Sirus

I do not consider centrifugal force to be a good way of explaining these things. What you said is all correct, but the idea of a force pointing outwards from the circular path of an object scares me. Understanding circular motion by way of a force that does not exist is not recommended, in my opinion.