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Ch4 HW4 S2008
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Ch4 HW4 S2008 (Homework)
KRISTIN SEILOFF
PHYS 2211 M & N Spring 08, section N05, Spring 2008
Instructor: Jennifer Curtis
Description
More on curving motion
Instructions
Reading: Sec. 4.13. Additional examples of curving motion are found in Sec.
4.18.
Web
Assign
Current Score:
76.5 out of 77
Due:
Wednesday, February 13, 200812:00 PM EST
1. 7/7 points  1/4 submissions
A child of mass
20
kg swings at the end of an elastic cord. At the bottom of the swing, the child's velocity is horizontal, and the speed is
12
m/s.
At this instant the cord is
3.60
m long.
(a) At this instant, what is the parallel component of the rate of change of the child's momentum?
= < 0
0
, 0
0
, 0
0
> (kg·m/s)/s
(b) At this instant, what is the perpendicular component of the rate of change of the child's momentum?
= < 0
0
, 800
800
, 0
0
> (kg·m/s)/s
(c) At this instant, what is the
net
force acting on the child?
net
= < 0
0
, 800
800
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Ch4 HW4 S2008
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, 0
0
> N
(d) What is the magnitude of the force that the elastic cord exerts on the child? (It helps to draw a diagram of the forces.)

due to cord
 = 996
996
N
(e) The relaxed length of the elastic cord is
3.56
m. What is the stiffness of the cord? (You will be given credit for this part if it is consistent with
your answer to part d, even if the answer to part d is incorrect.)
k
s
= 24900
24900
N/m
Solution or Explanation
(a) At this instant the magnitude of the momentum isn't changing, so the rate of change of the parallel component of momentum is zero.
(b) There is a rate of change of the perpendicular component of momentum, upward.
(c) The momentum principle says that the net force is equal in magnitude and direction to the rate of change of momentum, and the direction is
upward (+y), directed toward the center of the kissing circle. The x and z components of the net force are zero.
(d) If you draw a diagram of the forces, you'll see that the y component of the net force has two contributions, the upward force of magnitude F
of the elastic cord and the downward force of the Earth of magnitude mg. So the y component of the net force found in part (c) is equal to Fmg;
solve for F.
(e) Now that you know from part (c) the force exerted by the elastic cord, you can determine the stiffness from F = k
s
s, where s is the stretch
(change in length of the cord).
2. 6.5/7 points  4/4 submissions
A Ferris wheel is a vertical, circular amusement ride with radius
9
m. Riders sit on seats that swivel to remain
horizontal. The Ferris wheel rotates at a constant rate, going around once in
9
s. Consider a rider whose
mass is
58
kg.
At the bottom of the ride, what is the parallel component of the rate of change of the rider's
momentum?
= < 0
0
, 0
0
, 0 > kg·m/s/s
At the bottom of the ride, what is the perpendicular component of the rate of change of the rider's momentum?
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