Unformatted text preview: tells us that the boxcar must have moved This
such as Adobe Reader. ∆x = 3.32 m − 4.65 m = −1.33 m. Problem 10:
Let’s call the radius of the smaller disk r, and its mass m. Then the larger disk has a radius
of 2r and a mass of 4m (four times larger because the mass is proportional to area). Also,
the centers of the disks are a distance r + 2r = 3r apart.
Choose the center of the larger disk as the origin of our coordinate system. Then we can
calculate the center of mass location using
xcm = 3mr
3r
4m(0) + m(3r)
=
=.
4m + m
5m
5 With the numbers given, this works out to be 2.4 cm. Copyright c 2012 University of Georgia. Unauthorized duplication or distribution prohibited. Physics 1111
Spring 2012 PS #8 Solutions
Page 9 of 9 Problem 11:
The center of the ﬁrst meter stick will be halfway in between its endpoints; since it lies on the y axis, this is
(x1 , y1 ) = (0, 0.980). The second meter stick is on the
x, axis, so its midpoint is at (x2 , y2 ) = (0.620, 0). The
third meter stick’s midpoint is halfway between the points
(0.440, 0) and (0, 0.898), so (x3 , y3 ) = (0.220, 0.449). (We
can average in the x and y directions independently.)
Because the meter sticks are identical, they all have the
same mass. So we just average the positions of their centers
to obtain the position of the center of mass: y CM 1
1
xCM = (x1 + x2 + x3 ) = (0 + 0.620 + 0.220) = 0.280
3
3
1
1
yCM = (y1 + y2 + y3 ) = (0.980 + 0 + 0.449) = 0.476.
3
3 x So the center of mass is at (x, y ) = (0.280 m, 0.476 m). This document requires a JavaScriptenabled PDF reader
Problem 12:
such as Adobe Reader.
(a) The tree’s center of mass remains the same distance from the base
of the tree (i.e., the pivot point) as the tree falls. This means that
the center of mass will trace out a circular path as it falls (1/4
of a circle, to be precise). It does NOT follow a parabola! A
parabola is a very speciﬁc kind of curve associated with freefall.
Not every nonstraightline motion is parabolic. (b) Because the center of mass falls, and speeds up as it falls, there must be a vertical
component of total force pointing downward. The forces acting on the tree are the
force of gravity (downward) and the normal force due to the ground (with an upward
component). Since their combination points downward, the vertical component of the
contact force must be less than the tree’s weight. (It’s not zero, however, or else the
tree would be in freefall.) Copyright c 2012 University of Georgia. Unauthorized duplication or distribution prohibited....
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This note was uploaded on 08/15/2012 for the course PHYS 1111 taught by Professor Plascak during the Spring '08 term at UGA.
 Spring '08
 plascak
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