Download Introduction to Circular Motion

Introduction to Circular Motion
Newton's first law states that in the absence of a net force, an object will continue to
move in a straight line indefinitely at constant speed. However, a net force applied in
the direction of the velocity will cause an object to speed up and when applied in the
opposite direction, the object slows down. In these two cases, the net force causes a
change in the speed of the object. A third case is when the net force is perpendicular to
the velocity vector. In this case, the object continues at a constant speed but follows a
circular path. In such a case, the net force called the centripetal force is directed
towards the center of the circle. In accordance with Newton's second law, the
acceleration is also directed to the center of the circle and is appropriately called the
centripetal acceleration.
The formulas used in UCM (uniform circular motion) are:
acp = v2/r and Fcp = mv2/r
and as always units are important. The mass, m, must be in kg, the velocity, v, must be
in m/s, and the radius, r, must be in meters. We will restrict the motion of the object to
that of a horizontal circle, i.e. a circle parallel to the ground as on a merry-go-round.
What may not be intuitively obvious to the most casual observer is that circular motion
and rotary motion are not totally different from each other. Both are examples of
periodic motion where periodic motion is that type of motion that is repeated at equal
time intervals. The time needed for a complete interval is called the period. This
relationship can be expressed in the following formulas:
v = C/T = 2πr/T and a = 2πv/T
where C is the circumference of the circle in m and T is the time needed for one
revolution (period) measured in s.
The appropriate rotary quantities are expressed as:
s = rθ and v = rω
where s is the linear displacement measured in m, r is the radius in m, θ is the angular
displacement in radians (rad), and ω is the angular velocity in rad/s.
Circular Motion Problems
1) A 5.0 kg mass is swung in a horizontal circle. The radius of the circle is 2.0 m and
the period is π s.
(a) What is the time for one complete revolution?
The period is the time needed for one complete revolution, 3.14 s.
(b) What is the magnitude and direction of the velocity for the 5.0 kg mass?
m = 5.0 kg
r = 2.0 m
v = ΔC/ΔT = 2πr/T = 2*π*2.0 m/3.14 s = 4.0 m/s
The direction of the velocity is tangent to the circle at all points. If the string was to
break, the mass the mass would move in a straight line tangent to the circle. If
gravity is negligible then the net force would be zero and Newton's first law would
apply.
(c) What is the direction of the acceleration? Does the acceleration remain constant?
Explain your answer.
The direction of the acceleration is towards the center of the circle but only its
magnitude remains constant. Note that the direction of the acceleration vector
is constantly changing because it is always pointing in a different direction.
(d) What is the magnitude and direction of the centripetal force?
m = 5.0 kg
r = 2.0 m
v = 4.0 m/s
Fc = mv2/r
Fc = 5.0 kg*(4.0 m/s)2/2.0 m = 40. N
Because of Newton's 2nd law, the net force and the acceleration must be in the
same direction which in case is towards the center of the circle.
2) Apply Newton's 3rd law to Problem 1 above. When the string exerts a force of 40.N
on the 5.0 kg mass, the centripetal force can be considered as the action force.
(a) What is the reaction force?
The reaction force is the mass exerting radially outward on the string and is called
the centrifugal force.
(b) Why don't these two forces, the centripetal and the centrifugal force cancel out
leaving Fnet = 0? Justify your reasoning.
The centripetal and centrifugal force will never cancel out because they are not
acting on the same object. They are equal in magnitude and opposite in direction
but the centripetal force results from the string pulling inward on the mass and the
centrifugal force is the mass exerting an equal force radially outward on the string.
3) A watch has a second hand which is 2.2 cm long.
(a) Determine the speed of the tip of the second hand.
r = 2.2 cm
v = ΔC/ΔT = 2*π*r/T = 2*π*2.2 cm/60 s = 0.23 cm/s
In case you are wondering, the answer is NO! The velocity can be expressed in
either m/s or cm/s.
(b) Compare the velocity of the tip of the second hand as it passes over the 12 with
that of the 9.
If the watch is running properly, the speed or magnitude portion of the velocity
remains the same, 0.23 cm/s. However, the direction of the velocity as the
second hand passes 12, is straight down and to the right as it passes over the 9.