Download Guaranteed way of losing 7-8 points on exam (installment #2)

Circular Motion
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General Requirements
Special Case of Newton’s 2nd Law
Kinematics of circular motion
Dynamics of circular motion
Examples
Frequency and Period
More examples
Circular Motion – General Requirements
• What is criterion for circular motion?
• Must accelerate to side at certain rate!
• Rate determined by velocity, radius
Circular Motion and 2nd Law
• Circular motion is special case of 2nd Law
𝐹 = 𝑚𝑎
Forces that produce
that acceleration go
here
Required
acceleration
goes here
• Much like elevator (chapter 4)
𝐹 = 𝑚𝑎
FT
ma
Tension – weight
goes here
Elevator
acceleration
goes here
696
N
What is required acceleration?
• Two Isosceles Triangles
Radius and velocity triangles
Velocity triangle
Δr
v
r
r
• From similar triangles: Δv/v = Δr/r
• a = Δv/ Δt = v/r (Δr/ Δt) = v2/r
v
Δv
Δv = v Δr/r
Circular Motion Basics
• For circular motion a = v2/r
• Then right side 2nd Law must be mv2/r
𝐹 = 𝑚𝑎
Real forces that
provide this motion
go here
• If acceleration or force:
– Equal, stays in circle
– Less, spirals out of circle
– More, spirals into circle
Required for
circular
motion here
Circular Motion Caution
• “Reverse” type problem
• mv2/r is not a force, it is a requirement!
• a =v2/r goes on right side of F=ma, work
backward to get necessary forces
• Something else must supply actual force
– Tension
– Normal force
– Gravity
• Guaranteed way of losing 7-8 points on exam
(installment #2)
Again, remember
𝐹 = 𝑚𝑎
mv2/r does not go here
-
Goes here
Ftension + Fnormal + Fgravity + Ffriction = mv2/r
Example - Chapter 5 Problem 1
A child sitting 1.1 m from the center of a merry-go-round moves with a speed of 1.25 m/s. Calculate a) the
centripetal acceleration of the child and b) the net horizontal force exerted on the child (mass = 25 kg)
• Acceleration
𝑣2
1.25 𝑚/𝑠
𝑎𝑟 =
=
𝑟
1.1𝑚
2
= 1.42 𝑚/𝑠 2
• Required Force
𝑚𝑣 2 25 𝑘𝑔 ∗ 1.25 𝑚/𝑠
=
𝑟
1.1𝑚
2
= 35.5 𝑁
1.25 m/s
• What real force provides that force?
– Friction?
– Tension in string?
1.1 m
Example
•
A horizontal force of 210 N is exerted on a 2 kg discus as it rotates uniformly in a horizontal circle (at
arm’s length) of radius 0.90 m. Calculate the speed of the discus
• Available Force
𝑚𝑣 2 2 𝑘𝑔 ∗ 𝑣
=
𝑟
0.9𝑚
2
= 210 𝑁
• What velocity v will require that force?
• 𝑣=
0.9 𝑚∗210 𝑁
2 𝑘𝑔
= 9.7 𝑚 𝑠
=
189 𝑘𝑔 𝑚2 𝑠 2
2 𝑘𝑔
v m/s
0.9 m
Frequency and period
• Alternate way of describing velocity
• Frequency
–
–
–
–
Number of revolutions per unit time
Rpm, etc
always use seconds
v = 2πr f
• Period
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–
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Time to make one revolution
24 hour earth rotation
Always use seconds
v = 2πr /T
• Frequency Period reciprocals f = 1/T
2𝜋𝑟
Circular Motion Examples 1
• Example 5-1, 5-3
–
A 150-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.6 m.
The ball makes 2.0 rev/sec. What is the velocity and centripetal acceleration? What force
must be exerted on the ball to keep it revolving this way?
2𝜋𝑟
– v = 7.54 m/s, ar = 94.7 m/s2
– F = 14 N
• Example 5-2
–
What is the velocity of the moon going around the earth? What is the centripetal
acceleration? (r = 384,000 km = 384 x 106 m, T = 27.3 days = 2.36 x 106 s)
– 𝑣=
2𝜋𝑟
𝑇
– 𝑎=
𝑣2
𝑟
= 1022.35 𝑚/𝑠
𝑚
= .0027 𝑠2 = 2.78 ∙ 10−4 𝑔
Circular Motion Examples 2
• Problem 4
–
What is the acceleration of a spec of clay on the edge of a potter’s wheels turning at 45
rev/min if the wheel’s diameter is 35 cm?
2𝜋𝑟
– v=?
• Space station
–
What is the velocity and radial acceleration of the International Space Station orbiting at an
altitude of 400 km?
– r = 6.78 x 106 m (6.38 x 106 m + 400 km)
– T = 1.5 hours = 5400 s
– v = 7889 m/s = 28,400 km/h
– a = 9.19 m/s2
• Dachel on the slippery floor