Grade 12 Physics
Lesson #14 UCM Summary Notes
page #2
page #3
Short-Cut Formula for the magnitude of the Instantaneous Acceleration of UCM
We have derived |a|= v2 /r Memorize this!
where v is the speed and r is the radius
As we determined earlier, for UCM, Δv is always directed toward the center of the circle.
But a and Δv are always in the same direction. This is becausea = Δv/Δt , and division of a vector by a positive scalar always results in a new vector in the same direction as the original.
Therefore the direction of the instantaneous acceleration for UCM is toward the center of the circle.
Page #4
A New Name for the Instantaneous Acceleration of UCM
A Latin adjective is used by physicists to help them remember that the direction of the instantaneous acceleration for UCM is always directed “towards the centre of the circle”. Instantaneous UCM acceleration is called centripetal acceleration from the Latin petare = “to seek” and centri = “the centre”
The instantaneous acceleration for UCM will henceforth be referred to as “centripetal” acceleration with new symbol ac . The formula for the magnitude of the centripetal acceleration is now modified as: ac = v2 / r Memorize please!
Page #5
Other Formulas for UCM centripetal acceleration
We can use ac = v2 / r if we know the speed and radius of UCM. But suppose we don't know the speed, but know the time for the UCM to complete one revolution or rotation. This special time is called period or the time to complete one cycle and it has the symbol T. For example, the earth going around the sun is approximately UCM with a period of
T = 365.25 days.
We can quickly derive a formula foracin terms of r and T.
Page #6
Deriving a Centripetal Acceleration Formula in terms of T and r
The basic constant speed formula is
v = Δd/Δt Eq #1
Δd = 2πr Eq #2
Δt = T Eq #3
Without simplifying, sub equations #2 and #3 into #1:
v = 2πr/T Eq #4 We knowac = v2 / r Eq #5
Without simplifying, sub Eq #4 into Eq#5:
ac = (2πr/T)2 / r With pencil, simplify to a two-story ac = 4π2r / T2 Memorize please!
page #7
Deriving a Centripetal Acceleration Formula in terms of f and r
For UCM. sometimes we don't know v or T, but we know f , the number of revolutions or complete circles a particle goes through per unit time. (usually in units of seconds) fis called frequency. In SI, the unit of frequency is #cycles/s or 1/s or s-1 or the Hertz (Hz).
Frequency and period are reciprocals of each other. So f = 1/T Eq #6 But we know ac = 4π2r / T2
We write ac = 4π2ror = 4π2r (1/T2)
T2
ac = 4π2r f2 Memorize please!
Page #8
Review of the Circle Three for UCM
ac = v2 / r UCM circle equation #1
ac = 4π2r / T2 UCM circle equation #2
ac = 4π2r f2 UCM circle equation #3
The direction of acis always directed towards the centre of the circle.
If we want acceleration in units of m/s2, we must substitute MKS units into the formula.
Page #9
Example #1: The moon goes around the earth in approximately UCM every 27.3 days with respect to the stars. (called a sidereal month). If the centre-to-centre distance from earth to moon is 384,000 km, what is the moon's centripetal acceleration about the earth in m/s/s?
Given: T = 27.3 days X 24 h/day X 3600s/h = 2.36X106s
r = 384,000 km X 1000 m/km = 3.84X108s
Formula:Unknown: ac
ac = 4π2r / T2
Sub: ac = 4π2(3.84X108) / (2.36X106)2
ac = 2.72 X 10-3 m/s2 [toward the centre of the earth]
Note the moon is accelerating with a very tiny acceleration compared to projectiles near the earth's surface.
Page #10
Try Example #2: A car is moving on a circular track of radius 0.522 km. The magnitude of its centripetal acceleration is 4.00 m/s2. Find the speed of the car in km/h.
Given:
r = 0.522 km = 0.522 km X 1000 m/km =522 m
ac= 4.00 m/s2
Unknown: v =?
Formula: ac = v2 / r orv2 = r acor
v = (rac)1/2
Sub: v = (522 X 4.00)1/2
= 45.7 m/s X 3.6
= 165 km/h too fast!
Page #11
Example #3:The planet Mercury moves in an approximately circular path around the sun at an average distance of 5.8 X 1010 m, accelerating centripetally at 0.40 m/s2. What is the period of revolution about the sun in days?
Given: r = 5.8 X 1010 m ac= 0.40 m/s2
Unknown: T =?
Formula:ac = 4π2r / T2or
T = (4π2r /ac)1/2
Sub: T = (4π2 (5.8 X 1010 )/ 0.40 )1/2
= 7.6 X 106 s
=7.6 X 106 s X 1 h/3600 s X 1day/24 h
= 88 days
page #12
Example #4: A stone is whirled in UCM on a smooth sheet of ice. The stone traces out a circle of diameter 3.0 m as it accelerates centripetally at 93.0 m/s2. Find the frequency of rotation in Hertz and in rpm (revolutions per minute)
Given: r = 3.0 m / 2 = 1.5 m
ac = 93.0 m/s2
Unknown: f = ?
Formula: ac = 4π2r f2 or
f =(ac /(4π2r))1/2
Sub: f = ( 93.0 / (4π2 (1.5)))1/2
Answer: f = 1.25 Hz
Or 1.25 rev/s X 60 s/ min = 75 rpm