| Last revised 1999/11/02 |
#73. First, this is just one problem, not two, if you work it symbolically and plug in numbers last. Second, it help to compare it to a problem that asks for the acceleration that it takes to get each of two different blocks from rest to a set v. In part (b), consider the definition of torque.
#76. You know how to get an equation from the acceleration of each of the masses. Try to get another equation by considering the behavior of the pulley. What is the relation between the linear acceleration of the blocks and the angular acceleration of the pulley?
#85. There are hints already in the problem. For part (c), think about finding the easiest way to tackle the problem. The final speed of the rotor is intended to be 320 rev/min.
#87. You need 3 equations if you do not use time and 4 if you do. How can you avoid using time as a variable? The kinetic energy of the tower has two parts; what are they?
In practice, falling towers almost always break somewhere near the base before the top hits the ground.
#15. Compare the earlier problem (p. 179 #39) where a block was sliding on the loop-the-loop rather than a marble rolling. How is the setup of this problem different from the earlier one?
#33. The diagram is almost the entire proof. Can you find r sin (theta) on your diagram?
#41. Consider using x and y coordinates of both position and velocity.
#69. This problem is a form of the ballistic pendulum. You have one kind of physics while the particle is sliding, another for the collision, and a third while the rod and pendulum swing freely. [I don't mean that all three sets of principles are necessarily completely different.] Remember that "particle" is a code word for "having no significant size." Don't forget that during the collision there can briefly be large forces exerted on the rod at the pivot O . You are free to use any point as the origin of a calculation of angular momentum. How might you get rid of the effect of the large forces at O ?