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Phy 3221
Due: January 26, 2011
Homework set # 3
You are encouraged and welcome to work together on the homework assignments. But your
homework must be in your own handwriting.
If you expect credit for your homework, then it is best to make it easily readable.
Problem 1: Look at Figure 2-A on p 94. A block of mass m is released from a height
h = 5R above the ground and slides down the frictionless incline shaped as in the figure.
Give your answers in terms of m, g, and R. Assume that the size of the block is much
smaller than R. Be certain that the dimensions work out properly.
a. What is the force on the block at the bottom of the incline at point A?
b. At what speed does the block leave the track?
Textbook: Prob. 2-5
Textbook: Prob. 2-9
Textbook: Prob. 2-19 Assume that the object is launched off the top of a tall tower, where
x0 = y0 = 0 (for simplicity), and let the initial components of the velocity be v0x and v0y .
Don’t be concerned with the object hitting the ground. Here are some suggestions for this
challenging problem.
• Assume that you can solve it.
• The problem concerns a trajectory such that the distance from the starting point is
always increasing—that is, that the distance never reaches a maximum.
• You might wonder how this can make sense. If you throw a ball straight up then it
moves away from you for a while, and then it gets closer to you again. But if you
throw a ball horizontally away, then it will never get closer to you again. Somewhere
between vertical and horizontal is a special angle where the ball would get closer to
you if you just threw it at a slightly higher angle.
• So write down the usual formula for the position as a function of time for a trajectory.
To keep the algebra simple, let the object start at x = y = 0. Your expression should
allow for arbitrary initial components of the velocity, and you might call these v0x and
v0y .
• The distance is always supposed to increase, so the distance squared is also supposed
to always increase, so: Find an expression for the distance squared which depends
upon the time and depends upon the initial components of the velocity. Why do I
suggest the distance squared? It doesn’t involve a square root, and square roots just
complicate algebra.
• The distance squared is supposed to always increase. It is not allowed to have a
maximum. So, take the time derivative of the distance squared and find an algebraic
equation for the time when that derivative would be zero.
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• If this question has a reasonable answer (and it does), then you should be able to
find a condition on v0x and v0y so that there is no time when the distance reaches a
maximum. In other words, there should be a relationship between v0x and v0y such
that the algebraic equation that you just found has no solution for the time when the
distance squared is a maximum.
• The final answer to this problem is a maximum angle, and that will have a tangent
which is v0y /v0x , for the limiting case that the trajectory almost has a maximum
distance.