Download Lecture #4, June12

Lecture 4.4
Applications of Newton’s Second Law
Circular Motion
We have already talked about various applications of the Newton’s second law.
However, the only examples which we considered were examples of translational motion,
while studying kinematics we also learned how to describe rotational motion. Now we
shall look at rotational dynamics.
Think what you know about acceleration of the point-like object moving around a
circle.
When we discussed kinematics of rotation, we saw that rotation is always
accompanied with acceleration, known as centripetal acceleration. We proved that it is
always directed towards the center of the circle. Indeed in order to be able to make the turn
a particle should change the direction of its velocity towards the center of the arc path, so
the acceleration should be directed in a same way. The magnitude of this acceleration is
acp 
v2
.
r
(4.4.1)
1. Centripetal force
As we learned from studying the laws of Newtonian mechanics, acceleration can not
come from nowhere. If a body has acceleration then, according to the Newton's second
law, the nonzero net force should exist which produces this acceleration. Since we are
talking about centripetal acceleration, we shall call this force a centripetal force. It is
directed to the center of the circle and has the magnitude of
v2
F m .
r
(4.4.2)
I deliberately have not used any special letter for centripetal force, because there is no any
special centripetal force in nature. Indeed equation 4.4.2 is just another way of writing
Newton's second law for the case of the circular motion. This means that significance of
the force in the left-hand side of this equation is just the net force acting on the body. We
have already discussed that the net force is not a special force but just the result of the
vector addition of all the forces acting on the body. In a same way the centripetal force is
the result of the vector addition of all the forces acting on the body. You do not have to
show centripetal force as an extra force in the free body diagram. However, if the body
moves around a circle then it has centripetal acceleration directed to the center of this
circle. You have to take this into account when performing vector addition of the forces.
This is why it is convenient to choose one of the coordinate axes in the direction of the
radius of the circle and another in the direction perpendicular to the radius (tangential
direction).
Let us consider several simple examples. The passenger is sitting at the back seat of
the car which makes a turn. This passenger can move along the seat if he/she is not
holding or using the belt. This is because the passenger experiences the effect of inertia.
Indeed according to Newton's first law this passenger should continue to move along the
straight line if no forces are acting on him/her. In order to be able to make this turn he/she
must experience a force which will provide the centripetal acceleration. In the case of the
car it is either the force of the belt or the force of friction between the passenger and the
seat. As you can see, it is not some special centripetal force, but just one of the regular
forces acting on this person.
Another good example would be motion of the Moon around the Earth. Considering
Moon's orbit as a circle, we have to find the centripetal force which is responsible for its
circular motion. The only force acting on the Moon is gravitational force of the Earth. It is
directed to the center of the Moon's orbit. Even though it is centripetal force, but it has
common gravitational nature. If it is not this force acting on the Moon, it will just fly away
from the Earth along the straight line (according to the Newton's first law). Sometimes
people ask a question why the Moon is not falling down to the Earth if there is
gravitational attraction between them. This is because the Moon has finite orbital velocity.
In a sense it is falling, but it also moves in tangential direction, so finally it just makes one
more turn around the Earth. If something suddenly stops the Moon then it will fall down to
the Earth.
Example 4.4.1. The coefficient of static friction between the road and the tires of the
car is  . What speed will put the car on the verge of sliding as it rounds a leveled curve of
radius R?
When the car makes this turn it should have centripetal acceleration directed along
the ground to the center of the curve. The only force which may act on the car in this
direction is the force of friction. It is static friction, since the car does not slide from the
curve. So the force of static friction is the centripetal force for this case and we have
m
v2
 Ffs ,
R
The maximum value of this static friction force is Ffs max  N  mg . Since for the car is
on horizontal road then N  mg . This gives
2
vmax
m
 mg ,
R
2
vmax
 Rg ,
vmax  Rg ,
which is the maximum possible speed for this car before it slides from the curve.