Transcript Potential Energy - McMaster Physics and Astronomy

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J. P. Student
Multiple-choice answer sheets:
Physics 1D03
Centre of Mass
• Centre of Mass, Centre of Gravity
Serway 12.1-12.2; and parts of 9.5
Physics 1D03
Example:
The uniform beam (weight 400 N, length L) is supported by a pin
at one end, and a scale that can be placed anywhere along the
beam. What will the scale read if it is
scale
a)
b)
c)
d)
at the end of the beam (shown)?
at the centre of the beam?
at distance ¼ L from the pin?
at distance ¾ L from the pin?
Physics 1D03
Where should we choose the “pivot point” when calculating torques?
Answer: In Statics problems, it doesn’t matter.
Theorem: If the net force on a body is zero, then
the total torque due to all forces will not depend on
the choice of “pivot point”.
In particular, if Fnet = 0, then if tnet = 0 for one “pivot point”, tnet is also
zero for every other pivot point.
Physics 1D03
Centre of Mass, Centre of Gravity
Two particles on the x axis; total mass, M = m1 + m2 .
x2
x1
m1
m2
xCM
The centre of mass (CM) is defined as the location:
xCM
m1 x1  m2 x2

M
The centre of mass is an “average position” of the mass.
Physics 1D03
For many particles:
rCM
miri miri


mi
M
(Recall the position vector r has components x, y, z in R3.)
Symmetry:
The center of mass of any symmetric object lies on the
axis of symmetry and on any plane of symmetry.
For a uniform object, CM = center of geometry and mass
is distributed evenly around this point
Physics 1D03
Examples:
The CM is on the
symmetry axes.
The CM can be outside the object.
(It is on the line between the centres
of the two rectangles which make
up the “L”.)
2L/3
m
L/3
CM
The CM is closer to the more
massive object
2m
Physics 1D03
Example:
Three particles m1=m2=1.0kg and m3=2kg are located like so:
y(m)
2
1
m3
rCM
m1
1
m2
2
3
x(m)
What is the center of mass of the system ?
Physics 1D03
Quiz: Baseball bat
A baseball bat is sawn in half at its
centre of mass. Which piece is
heavier?
A) The short piece
B) The long piece
C) Both pieces have the same mass.
Physics 1D03
• For an object, like the baseball bat, torques due to the
gravitational forces are zero around the center of mass
• But, this is not the point around which mass is evenly
distributed b/c the object has a non-uniform mass distribution
• The lighter side of the bat has a larger moment of inertia,
since it is longer
Physics 1D03
Centre of Gravity
The CM is also the location of the centre of gravity. When we
consider the rotational equilibrium of a rigid body, we can treat the
gravitational force as if it were a single force applied at the centre of
mass. A suspended object will hang with its CM vertically below the
point of suspension.
CM
mg
Physics 1D03
Quiz: A hemisphere on a ramp.
A uniform solid hemisphere is
placed on a ramp. Which of the
pictures shows how it rests in
equilibrium?
A) The top picture
B) The middle picture
C) The bottom picture
Physics 1D03
Centre of Gravity
Why can we place the gravitational force at the CM?
Calculate the torque (about O)
due to the three weights on the
x axis:
x3
O x2
x1
t  x1m1 g  x2 m2 g  x3 m3 g
m1g
Now calculate the torque as if a
single, total weight were placed
at the CM:
O
m2g
m3g
xCM
t  xCM Mg
x1m1  x2 m2  x3 m3
but xCM 
M
Mg
The two methods give the same torque !
Physics 1D03
Replacing the real gravitational forces by a single
force, equal to the total weight, and placed at the
centre of gravity (which may or may not have any
mass at this point), will not change the total
gravitational torque (about any pivot).
This means that the external forces needed to hold a rigid body
in equilibrium can be calculated as if gravity were a single force
applied at the centre of gravity.
In a uniform gravitational field, the centre of gravity is at the
centre of mass (eg: a sphere).
Physics 1D03
?
Example: how far can a pile of bricks
lean without falling over? Is it possible
for the top brick to be entirely past the
edge of the table?
Physics 1D03
Example: how far can a pile of bricks lean without falling over?
If the centre of gravity of the entire pile is
past the edge of the table, there will be a
clockwise torque about the edge of the table
(point E), and the whole stack will tip (rotate
clockwise) about E.
E
If the centre of gravity of the pile is not past
the edge of the table, the gravitational torque
(about E) will be counterclockwise, and the
stack will be stable (at least it will not tip at E).
The 4th brick can be entirely
past the edge
Physics 1D03
Summary:
Center of mass:
rCM 
miri miri

mi
M
Gravitational forces may be replaced by a single force
(weight) located at the CM
For a uniform object, CM = center of geometry and mass
is distributed evenly around this point
Physics 1D03