Course
MA1241
& MA1242 - Classical Mechanics I & II
Questions
and
Answers (2008-09)
Q: Angular momentum and torque.
Could you please explain why the
angular
momentum vector and torque vector of a rotating body are
perpendicular
to the direction of rotation, i.e. why L = r X P and T = r X F.
A: The
expressions you
write are valid for a point particle. Angular momentum and torque
for a
point-particle located at r
and moving with momentum P
under the action of a force F are
simply
defined in this way. You should take this definition as
such. However, there is a reason for choosing this particular
definition. The reason is that L
and T defined in this
way are
useful. This is because
they
satisfy the equation dL/dt = T, which we have used in many
examples.
From the definition it
follows that the two vectors L
and T are perpendicular
to
the position vector, r. In
the
case of a point particle, the angular momentum is perpendicular to
the plane identified by the two vectors r and v.
If this plane is fixed both the
angular momentum and the
angular
velocity of the particle are parallel
to
each other and perpendicular to the plane of r and v (I suppose that this what
you
mean).
Notice, however, that the angular momentum of a rotating body is
not
always parallel to its angular velocity. Some sentences in chapter
6 of
the book may be a bit misleading. In that chapter (for instance in
section 6.4) only the component of the angular momentum of
the body in the direction (z) about which the body rotates is
computed. L as a vector
will
have in general other non-zero components. In fact we have seen
this
explicitly in the case of the rotating skew rod (see examples in
section 7.2 of the book): the system has angular velocity along
the z direction, but
its angular
momentum is always perpendicular to the rod, forming a fixed
non-zero
angle with the direction of the angular velocity vector.
Q:
Direction of friction.
In the solution for question 5 you have a friction force which
is
directed down the slope that the car is moving on. And since
friction
opposes the motion that would occur in its absence, my question
is:
What causes this motion up the slope that friction is directed
against?
As I can only see forces down the slope.
A: In some
cases, and
this is one such case, it can be a bit tricky to determine which
way
the friction force is directed. One way of thinking about the
problem
is the following. Consider what happens without friction. In that
case
there is only one value of the angle alpha for which the car does
not
skid. This angle is determined by the condition that the resultant
of
the only two forces present (weight and normal force, N, exerted by the
road) be equal to M (the
mass
of the car) times the centripetal
acceleration. Let's call alpha_0 this angle at which the car does
not
skid even in the absence of friction. If (still with no friction)
you
decrease alpha making it smaller than alpha_0, the car will skid
outwards because you have decreased the component of N in the
centripetal direction. If you do the same (decreasing alpha) in
the
presence of friction, this additional force will oppose the
tendency of
the car
skid outwards and so it will point downwards along the road. Now
consider what happens if you increase alpha making it greater than
alpha_0. Again consider first the situation without friction. As
alpha
becomes greater than alpha_0 the centripetal component of N grows.
Therefore the resultant of the vector sum of N and weight causes the
car to slide downwards. If you now turn on friction in this case,
the
effect will be to oppose this motion downwards. Therefore in this
case
the direction of friction is up along the road.
The motion up the slope you refer to is due to the fact that the
car is
moving along a circular trajectory with velocity v. This means that it
has a
centripetal acceleration equal to v^2/R, which must be
provided by the forces acting on the car, i.e. weight and normal
force. When
the angle alpha is decreased the centripetal
component
of the resultant force on the car decreases and the vertical
component
increases accordingly. This means that the force in the horizontal
direction is not capable of producing the centripetal acceleration
anymore. The car therefore skids outwards, which means up the
slope.
A perhaps more intuitive picture is provided by the point of view
of an
observer moving with the car. In this case the motion up the slope
is
due to a fictitious force, the cetrifugal force. When alpha is
varied
the balance between real and fictitious forces is lost and the car
skids.
This note
provides details of the explanation given above from the point of
view
of an inertial reference frame.
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