Download 5.3 Centripetal Force

» From Newton’s 2nd Law we know
that whenever an object accelerates,
there must be a net force acting on
the object to create the acceleration.
» The net force causing centripetal
acceleration is called centripetal
force, and points towards the center
of the circle.
Fc
Fc
Fc
nd
2
»Newston’s
Law say ΣF=ma,
so…
»ΣFradially= Fc = mac
2
»Fc =
𝑣
m( )
𝑟
The sum of the forces acting in the radial
direction equals the mass times the centripetal
acceleration
» “Centripetal force” is not a new or separate
force!
» “Centripetal force” describes (or labels) the
net force pointing toward the center of a
circular path.
» It is the vector sum of all forces pointing
along the radial direction.
» The centripetal force could be due to
tension, or friction, or lift, etc. A FBD will
help to identify which force(s) is(are)
causing centripetal acceleration.
» A model airplane has a mass of 0.90kg and
moves at a constant speed on a circle that is
parallel to the ground. The path of the airplane
and its guideline lie in the same horizontal plane
because the weight of the plane is balanced by
the lift generated by its wings. Find the tension
in the guideline for speed of 19m/s and 38m/s.
The length of the guideline is 17m.
Speed = 19m/s; T=19N
Speed=38m/s; T=76N
» When a car moves without skidding
around a curve, what provides the
centripetal force to keep the car on the
road?
» static friction between the road and the
tires provides the centripetal force to
keep the car on the road.
» When a car moves at a steady rate
around an unbanked curve, the
centripetal force keeping the car on the
road comes from the static friction
between the road and the tires.
» Some free body diagrams for a car going
around a curve would look like this:
» EX 7: At what maximum speed can a car
safely negotiate a horizontal unbanked turn
(radius = 51 m) in dry weather (coefficient
of static friction = 0.95) and again in icy
weather (coeff of static friction = 0.10)?
» Ff = μ N = μ mg
» Fc =
𝑚 𝑣2
𝑟
𝑚 𝑣2
𝑟
» Fc = Ff
so
= μ mg
» Notice
that the mass will cancel out of the equations above:
2
𝑚𝑣
= μ mg
𝑟
» On dry road: v = 51 ∗ 0.95 ∗ 9.8 = 21.6 m/s
» On icy road: v = 51 ∗ 0.10 ∗ 9.8 = 7.1 m/s
» Kevin is playing on a merry-go-round in a
park. He is hanging on to the railing at the
edge of the merry-go-round, 1.5 m from
the center.
» a) If Kevin has a mass of 65 kg, and the
railing pulls him in with 260 N, how fast is
he moving?
» b) How long will it take him to make one
complete revolution around the merry-goround? (Hint: How far does he have to
travel to go around the circle?)
» You go on the floor drop ride in a local
amusement park. In the ride, you stand
along the walls of a giant drum (5 m radius),
which is rotated at fairly high speed. Once
you are spinning at full speed, the floor is
dropped out from underneath you, but you
don’t fall with it. If the coefficient of friction
between you and the wall is 0.8, how fast
does the drum need to be spinning for you
to stay up when the floor drops? Guess your
mass as best you can to use in the
calculation. (Hint: Think about what force
makes you spin in a circle with the drum and
how that is connected to the friction.)
Answer: 7.83 m/s
» When an airplane or stunt driver completes
a loop de loop, they often vary their speed as
they complete the circular turn (non-uniform
circular motion). Nonetheless, they can
maintain a constant speed and complete the
turn. Let's assume uniform circular motion
to examine the FBD at four locations on the
loop. The propulsion and braking forces are
omitted because they are not acting in the
radial direction.
» In every case, the sum of the Radial forces
must =
𝐹=
mv2
r
If the mass of the car is
1100kg, and the radius of
the loop is 10m, what is the
minimum speed that the
car must maintain to make
it around the loop?
»P. 156 #13-17, 21, 23