Week 3

Sample problems:

1) A block is on an inclined plane whose angle theta is slowly steepening. The coefficient of static friction is 0.5, and the coefficient of kinetic friction is 0.25. Find the critical angle where the block starts to slide, and also find its acceleration when it starts to slide.

2) A cannon ball is shot at a given angle theta, and it goes a distance D, ignoring air resistance. What happens to its range if the ball is shot with the same horizontal velocity, but double the vertical velocity?

Solutions:
1) At the critical angle, the static friction is 0.5 times the normal force. The normal force is mg*cos(theta). The static friction force must balance the component of gravity along the incline, which is mg sin(theta), so we can say: mg*sin(theta) = 0.5 mg*cos(theta) and thus tan(theta) = 0.5, so theta = arctan(0.5). When it starts to slide, its acceleration is along the incline, and the net force in that direction is mg*sin(theta)-0.25*mg*cos(theta). This means the acceleration is a = g*sin(theta)-0.25*g*cos(theta), where theta is the solution to the above.
2) Double the vertical v means the time it takes the acceleration of gravity to take that velocity to zero goes up by 2, so the time in the air goes up by a factor of 2, so the distance travelled horizontally goes up by a factor of 2.

Circular Motion and Use of Energy
For motion in a circle of radius r at constant speed v, the key thing to recognize is that this implies an acceleration of magnitude v^2/r, pointed toward the center of the circle. This also implies if the object has mass m, the net force on the object needs to be mv^2/r, regardless of what the sources of the force is. Whatever those sources are, they must add up to mv^2/r, or the object cannot be moving in a circle. The classic example of this is if there is some kind of external constraint forcing the motion in a circle, like a rock on a string, or a car being driven on a circular road. In the first case, the string is called upon to provide whatever force it takes to maintain its length, for that is one strings do. In the second case, it is static friction with the rolling tire that yields whatever force is needed to keep the car on the circular road.

The Work-Energy Theorem
The kinematics formula that tells us that the change in v^2 is equal to 2ax can be recast in a form known as the work-energy theorem. All we do is multiply through by m/2, so we get the change in 1/2 mv^2 equals max. If we use F = ma to replace ma with F, we get that the change in 1/2 mv^2 = Fx. This motivates that we give these quantities names and interpret them as important physical entities: the first we call kinetic energy, and the second we call work. Now we view the equation as a way to track changes in kinetic energy by tracking the work done by some force F applied over a distance x, where F must point in the same direction as x.

Sample problems:
3) I swing at speed v a ball of mass m at the end of a rope of length r. Neglect gravity. What is the tension in the rope? Now include gravity and let the rope slant down at an angle theta from the horizontal. Find theta, and find the new tension in the rope.

4) A roller coaster starts at height h, drops to the ground, then rises up to height y, which is less than h. What is the speed of the roller coaster ride when it gets to y, if the track is frictionless?

Solutions:
3) T = mv^2/r if no gravity, and if there is gravity, it is T*cos(theta) that equals mv^2/r. The vertical problem tells us that T*sin(theta) = mg, so taking the ratio of the second equation to the first equation gives tan(theta) = gr/v^2, or theta = arctan(gr/v^2). Squaring the two equations and adding them, using sin^2 + cos^2 = 1, gives us T^2 = m^2*v^4/r^2 + m^2*g^2, so T is the square root of that. (Alternatively, find the components of T and use the Pythagorean theorem to get T.)
4) Equate the kinetic energy, 1/2 * mv^2, to mg times the change in height, so 1/2 * mv^2 = mg*(h-y), and solve for v^2 = 2g*(h-y), which is the same as you'd get in a free fall from h to y. Taking the square root of both sides gives v.