Transcript 7.1 Circular Motion

Circular
Motion
Uniform Circular Motion
An object moving uniformly (at the same speed) in a circle is
constantly changing direction which means it is constantly
changing its velocity which means it is constantly
accelerating.
Acceleration With Uniform Motion
In the diagram, v2 + (– v1) = ∆v. The triangle formed by v1 , v2
and ∆v is similar to the triangle formed by r1, r2 and d. As the
angle θ approaches zero, ∆v is at a right angle to the then
coinciding sides v1 and v2 . Thus the change in velocity is
directed towards the centre of the circle.
Instantaneous
velocity
Δv
change in
velocity
Centripetal Acceleration
Since the change in velocity has a direction towards the
circle’s centre, the acceleration also is directed towards the
circle’s centre because a = ∆v/t. The name given to the
acceleration of an object moving in a circle is centripetal
acceleration, “centre-seeking acceleration”.
Centripetal Acceleration Formula
In the diagram (a), triangle ABC is similar to the triangle in
(b). This means that Δv/Δl = v/r (equation 1). Rearranging
this expression gives Δv = vΔl/r (equation 2). Acceleration is
defined as a = Δv/Δt (equation 3). Substituting the Δv in eq.3
with its equivalent in eq. 2 produces a = vΔl/rΔt. But Δl/Δt is
linear speed, v of the object: v = Δl/Δt. So, a = vv/r = v2/r.
The acceleration of an object (centripetal acceleration) in
uniform circular motion is directed towards the centre at:
Circular Velocity and Acceleration
Circular velocity is always a tangent to the circle of motion
but circular acceleration is always directed towards the
centre of the circle of motion. When its accelerating force is
removed, an object flies off at a tangent to its former motion.
The Perpendicularity of Circular Velocity and Acceleration
An object moving in circular motion constantly has its
velocity in a direction perpendicular to its acceleration.
Frequency, Period and Velocity in Circular Motion
The frequency (f) is the number of revolutions per unit time
and is the reciprocal of period (T) which is the time for one
revolution ( f = 1/T ). Circular speed (v) is v = d/t. Since the
distance travelled is the circumference of the circle,
v = 2πr/T.
Acceleration, Frequency, Period and Velocity in Circular Motion
From a = v2/r, substitute v = 2πr/T into a = v2/r. So now
a = v2/r becomes a = (2πr/T)2/r . This is a = 4π2r2/T2r . A r in
the numerator cancels with r in denominator to give
a = 4π2r/T2 .
Problem 1: A Revolving Ball’s Accel;eration
A 150 g ball at the end of a string is revolving uniformly,
horizontally with a radius of .600 m. The ball has a frequency
of 2.00 Hz (makes 2.00 revs/s). What is its centripetal
acceleration? Thus T = 1/2.00 Hz = .500s .
F = 2.00 Hz
R = .600 m
v = 2πr/T = 2(3.14)(0.600m)/.500s = 7.54 m/s
The a = (7.54 m/s)2/0.600 m = 94.8 m/s2 .
Problem 2: The Moon’s Centripetal Acceleration
The moon has an average orbital radius of 384,000 km and a
period of 27.3 days (T). Determine the acceleration of the
moon towards the earth.
a = v2/r = (2πr)2/T2r = [2(3.14)(3.84E8 m)]2/(2.36E6 s)2(3.84E8)
a = .00272 m/s2 = 2.72E-3 m/s2 . Comparing this to g at earth:
2.72E-3 m/s2(1 g/9.80 m/s2) = 2.78E-4 (Fraction of g at earth)
Centripetal Force
From Newton’s Second Law, the sum of the forces on an
object (unbalanced force) equals the product of the mass and
the acceleration of the object (ΣF = ma) . Thus the centripetal
force on a object with circular motion is Fc = mv2/R . This
force’s direction is towards the centre of the circular motion.
Circular Motion Requires A Constant Centripetal Force
Without a constant force towards the centre of circular
motion, the motion becomes linear motion, motion in a
straight line.
There is no Centrifugal Force
A common misconception is that there is an outward, socalled, centrifugal (“centre-fleeing”) force. Consider a ball
circling on a string. For the action force of the string on the
ball, there is an outward reaction force of the ball on the
string but not on the ball itself. If there were a centrifugal
force, a ball would fly outward from a tangent when released
from its circular motion.
If centrifugal was
real
Actual tangential
motion after release
The Cause of the Apparent Centrifugal Force
The supposed outward force felt is a result of an object’s
inertia, its tendency to “want” to move in a straight line.
Problem 3
Calculate the centripetal force acting on a 1.5 kg object
whirling at a speed of of 2.3 m/s in a horizontal circle of
radius 0.60 m.
Fc = mv2/r
Fc = (1.5 kg)(2.3 m/s)2/0.60m
Fc = 13 N
Problem 4
A car travelling at 14 m/s goes around an unbanked curve in
the road that has a radius of 96 m. What is the centripetal
acceleration?
ac = v2/r = (14 m/s)2/96 m = 2.0 m/s2
Problem 5
A plane makes a complete circle with a radius of 3,622 m in
2.10 min. What is the speed of the plane?
V = 2πr/T = (2π)(3,622 m)/(2.10 min)(60 s/min) = 181 m/s
The Forces on a Tetherball
The tension force in the rope has components that balance
the weight and supply a centripetal force.
Tension
force in
rope
Weight =
mg
Tension component balancing weight
Tension component
causing centripetal force
Vertical Circular Motion
At the top position, the centripetal force equals the tension
force (T) plus the weight (Fg = ma). Fc = T + Fg
At the bottom position, the centripetal force equals the
tension force minus the weight. Fc = T – Fg
Minimum Speed to Maintain Circular Motion at Top
At the top, FT + mg = mv2/r (centripetal force).
When FT = 0 (tension in rope is zero), circular motion stops.
Thus the minimum speed at the top needed to continue
circular motion is mg = mv2/r which simplifies to g = v2/r and
v2 = gr and v = √(gr).
A 0.150 kg ball on a 1.10 m cord, swung in a vertical circle
would take a minimum speed of 3.28 m/s to keep circling.
(v = √(9.80 m/s2)(1.10 m) = 3.28 m/s). Note mass has no effect
The Cord Tension at the Bottom of a Vertical Motion
A 0.150 kg ball is swung in a vertical circle in a 1.10 m radius.
What would be the tension on the cord at the bottom of its
motion if it is swung at a speed of 6.56 m/s?
At the bottom FT – mg = mv2/r ; so, FT = mg + mv2/r
FT = (0.150 kg)(9.80 m/s2) + (0.150 kg)(6.56 m/s)2/1.10 m =
7.34 N
The Cord Tension at the Top of a Vertical Motion
A 0.150 kg ball is swung in a vertical circle in a 1.10 m radius.
What would be the tension on the cord at the top of its motion
if it is swung at a speed of 6.56 m/s?
At the top FT + mg = mv2/r ; so, FT = mv2/r - mg
FT = (0.150 kg)(6.56 m/s)2/1.10 m - (0.150 kg)(9.80 m/s2) =
5.87 N
Seat Force in a Ferris Wheel
At the top of a Ferris wheel, the seat force up and weight
force down add to the centripetal force which is down.
At the bottom of a Ferris wheel, the seat force up and the
weight force down add up to the centripetal force which is up.
Thus a Ferris wheel seat exerts less force on a person at the
top of its motion and more force on a person at the bottom of
its motion.
Centripetal
force
Weight or
gravity
Weight or
gravity
Centripetal
force
Seat force
Seat Force
Speed Breaking a String With Vertical Circular Motion
A string requires a 135 N force to break it. A 2.00 kg mass is
tied to this string and whirled in a vertical circle with a radius
of 1.10 m. What is the maximum speed that this mass can be
whirled without breaking the string?
Centripetal force
Tension =
135 N
Weight = mg
At the bottom of the motion, the tension force is the greatest.
At the bottom FT – mg = mv2/r ; so, FT = mg + mv2/r
135 N = (2.00 kg)(9.80 m/s2) + (2.00 kg)(x m/s)2/1.10 m
135 N = 19.6 N + 1.8181(x m/s)2
x2 = (135 - 19.6) kgm/s2 / 1.8181 kg = 63.4729 m2/s2
x = √(63.4729) = 7.97 m/s
Tension Force From Period
What is the tension force at the top of a vertical circular
motion for a 1.7 kg object moving at the end of a 0.60 m
string that takes 1.1 s for a revolution?
Centripetal force
Weight = mg
Tension force
At the top FT + mg = mv2/r ; so, FT = mv2/r – mg
Recall that v = 2πr/T so,
FT = (1.7 kg)(2π(0.60m)/1.1 s)2/0.60 m - (1.7 kg)(9.80 m/s2) =
FT = 33.279 N – 16.66 = 16.619 N = 17 N
Force Exerted on Curved Road Surface
A 826 kg vehicle travelling at 14.0 m/s goes over a hill with a
61.0 m radius of curvature. What is the force exerted on the
Centripetal force
road by the car at the crest of the hill?
Car force is down but
equal to the normal
force
Weight = mg
At the top of the hill, Fcar = mv2/r – mg , since Fc = FN + mg.
Thus Fcar = (826 kg)(14.0 m/s)2 / 61.0 m – (826 kg)(9.8 m/s2)
Fcar = 2654.03 N – 8094.8 N = -5440.77 N or 5 440 N down
Supplying Centripetal Force for Vehicles on Curves
For a vehicle to make a curve, it must have an inward
centripetal force. On a flat road, this force is supplied by the
road reaction force to the friction force applied by the
vehicle’s tires on the road.
When Is Friction too Low to Keep a Vehicle Turning?
A 1000 kg car rounds a curve on a flat road of radius 50 m at a
speed of 50 km/h (14 m/s). Will the car make the turn if the
pavement is dry and the coefficient of static friction is µ = 0.60?
F = mv2/r = (1000 kg)(14 m/s)2/(50 m) = 3900 N
The normal force equals the weight of the
vehicle = (9.80 m/s2)(1000 kg) = 9800 N .
Ffr = µFN = (0.60)(9800 N) = 5900 N
Since the friction force is much greater than the
required centripetal force, the centripetal force is easily supplied
by friction and the car turns the curve with no issues.
When Is Friction too Low to Keep a Vehicle Turning?
A 1000 kg car rounds a curve on a flat road of radius 50 m at a
speed of 50 km/h (14 m/s). Will the car make the turn if the
pavement is icy and the coefficient of static friction is µ = 0.20?
F = mv2/r = (1000 kg)(14 m/s)2/(50 m) = 3900 N
The normal force equals the weight of the
vehicle = (9.80 m/s2)(1000 kg) = 9800 N .
Ffr = µFN = (0.20)(9800 N) = 2000 N
Since the friction force is less than the
required centripetal force, the centripetal force can not be supplied
by friction and the car will skid.
Banked Curves
Banking or angling
curves creates a force
component inward that
can act as the centripetal
force to keep a vehicle
moving around a curve,
even without friction.
This inward-acting force
is a component of the
normal force (reaction of
road to vehicle weight)
and can be determined
by Nsinθ. The angle θ
between the normal and
vertical is the same as
the angle which the road
is banked at.
Friction Force on a Banked Curve for a Still Vehicle
For a non-moving vehicle on a banked curve, the friction
force of the tires on the road acts in a direction up the incline
to oppose the motion tendency to slide down the incline.
Friction Force on a Banked Curve for a Moving Vehicle
For a moving vehicle on a banked curve, the friction force of
the tires on the road acts in a direction down the incline to
oppose the motion tendency of the moving car to slide up the
incline due to its inertia.
Forces Providing The Centripetal Force For Turning
Together, the inward force from the normal force and the
inward force from the friction force provide the centripetal
force for turning a vehicle.
Friction, F
Component of friction
adding to centripetal force
Fy = Fcos θ , F = μFN
θ
Angle, θ
Formula For Banked Curves
The normal force plus the weight equals the centripetal force.
From the force diagram, tanθ = Fc / mg
tan θ = mv2/r / mg = v2/rg
tanθ = v2/rg
Normal
Note that tan θ
is rise/run or slope
θ
Fc
Weight = mg
θ
Banked Curve Problem
At what angle should a frictionless curve be banked if a car is
to safely round the curve of radius 475 m at a speed of 79
km/h (22 m/s)?
tan θ = v2/rg = (22m/s)2/(475 m)(9.80 m/s)
tan θ = 0.10 ; Θ = tan-1(0.10) = 5.9o
Banked Curve Problem
A car is rounding a 515 m frictionless curve. If the curve is
banked at an angle of 12.0 degrees, what is the correct speed
for the car?
tan θ = v2/rg
v = √(rgtan θ) = √[(515 m)(9.80 m/s2)(tan 12.0)] = 32.8 m/s
v = 32.8 m/s (118 km/h)
Summary of Banked Curve Relationships
A
AQuestion
a highway curve that has a radius of curvature of 100 meters is
banked at an angle of 15 degrees. What is the vehicle speed
which this curve is appropriate if there is no friction between the
road and the tires? Also, on a dry day, when friction is present, the
automobile negotiates the curve at a speed of 25 meters per
second. What is the minimum value of the coefficient of friction
that will keep the car from siding?
Factors Affecting Tension and Centripetal Forces
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A
AAnswer
Banked CurveIn the first case let’s assume that the curve is frictionless [much like the conditions that might exist during a freezing rain]. As
always, the first step will be to make the freebody diagram describing all of the relevant forces acting on the car as it passes through the curve. In
this case the gravitational force Fg is straight down, while the normal force FN is directed up and toward the left, perpendicular to the roadway.
The acceleration is centripetal and is directed left, toward the center of the curve. Here I want to show you a slightly different way of solving a
Newton’s 2nd Law problem.
Since this is a Newton’s 2nd Law problem the sum of the forces in the problem must be equal to the product of the mass and the acceleration.
If you make a vector diagram, attached, adding the two forces acting on the car [Fg and FN], the sum of these two vectors should be equal to the
mass times the acceleration. The acceleration here is centripetal and so is directed toward the left, while the gravitational force is directed down.
Therefore, the resulting triangle is a right triangle and so these forces can be related using sin, cosine or tangent.
Using this triangle, we can determine the critical velocity of the car as it passes through this banked, frictionless curve using
tan(Q) = m*ac/Fg = [m*v2/R]/[m*g] = v2/[R*g]
v = sqrt[R*g*tan(Q)]
What if there IS friction? What would be the maximum safe speed for a car to pass through this same curve if the coefficient of friction between
the car and the road is and the curve is still banked at an angle ? Now the freebody diagram also includes the frictional force Ff directed
down and parallel to the incline.
Now, the resulting vector diagram includes three forces adding up, tip to tail, to the resulting mass times the acceleration.
As before, the acceleration is centripetal and is directed toward the left. The added frictional force Ff is perpendicular to the normal force FN.
Since the frictional force is parallel to the surface that provides it, the frictional force meets the horizontal at the same angle Q as the incline.
The normal force, therefore, meets the vertical at the same angle Q, since the normal force is perpendicular to the frictional force.
The problem can now be completed by resolving the normal force FN and the frictional force Ff into components that are either parallel to or
perpendicular to the direction of the centripetal acceleration.
Having broken the frictional force Ff and the normal force FN into their horizontal and vertical components, the vectors in the horizontal direction
can be added up to equal the mass times the acceleration:
FN*sin(Q) + Ff*cos(Q) = m*ac
While the forces in the vertical direction can be made equal to one another since there is no acceleration in the that direction.
FN*cos(Q) = Fg + Ff*sin(Q) = m*g + FN*mu*sin(Q)
Typically, in a problem of this type there will be two unknown quantities, the normal force FN and the maximum safe velocity v.
If you solve each equation above for the normal force FN
FN = [m*v2/R]/[sin(Q)+mu*cos(Q)
FN = [m*g]/[cos(Q)- mu*sin(Q)]
and then make the two equations equal to one another, it is relatively easy to solve for the maximum safe speed v.
[m*v2/R]/[sin(Q) + mu*cos(Q)] = [m*g]/[cos(Q) - mu*sin(Q)]
Solve for v = sqrt{[(R*g*(sin(Q)+mu*sin(Q)]/[cos(Q)-mu*sin(Q)]}
A
The relationship between maximum speed, radius and the bank
angle can be found from considering the forces
FNsinθ>mv2r
FNcosθ=mg
mgsinθcosθ=mv2r
tanθ=v2rg
or with friction
FNsinθ+μFNcosθ=mv2r
FNcosθ−μFNsinθ=mg
sinθ+μcosθcosθ−μsinθ=v2rg
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