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Uniform Circular Motion

Have you ever ridden on the amusement park ride shown below? As it spins you feel as though you are being pressed tightly against the wall. The ride then begins to tilt but you remain “glued” to the wall. What is unique about moving in a circle that allows you to apparently defy gravity? What causes people on the ride to “stick” to the wall?


Amusement park rides are excellent examples of circular motion. When an object is moving in a circle of constant radius and its speed is constant, it is moving with uniform circular motion. When objects are moving in a circular path, speed is constant but direction continuously changes as it moves along the circle. Therefore they are experiencing centripetal acceleration towards the centre of the path. Centripetal is Latin for “centre-seeking”.


For example, consider an object as it moves

from point P to point Q as shown. If its

velocity changes from v1 to v2 then: ∆𝑣 = 𝑣2 − 𝑣1

Using triangle congruencies and the

equations v = ∆d/∆t for the distance travelled and

a = ∆v/∆t then as ∆t approaches zero yields:

𝒂𝒄 = 𝒗𝟐

𝒓

Note that since v1 and v2 are perpendicular to the

radii of the circle, the change in velocity and

acceleration vector points directly towards the

centre of the circle. Acceleration that is directed

towards the centre of a circular path is called

centripetal acceleration . This is the instantaneous

acceleration towards the circle centre.

Centripetal Acceleration

SUMMARY of UNIFORM CIRCULAR MOTION

occurs when an object moves in a circle of constant radius and its speed is constant

since direction changes the object experiences acceleration which is always directed toward the centre of the circle

𝒂𝒄 = 𝒗𝟐

𝒓

where ac is the centripetal acceleration (m/s2) v is the velocity (m/s) r is radius of the circular path (m) Note: ac is the instantaneous acceleration (∆t is very small)


Example #1: A child rides a carousel with a radius of 5.1 m that rotates with a constant speed of 2.2 m/s. Calculate the magnitude of the centripetal acceleration of the child.

ac = 𝒗𝟐

𝒓

ac = (𝟐.𝟐

𝒎

𝒔)𝟐

𝟓.𝟏 𝒎

∴ ac = 0.95 m/s2


Sometimes the speed of an object moving with uniform circular

motion is unknown. Often we can measure the time it takes for the object to move once around the circle, or the period (T).

If the object is moving too quickly, you would measure the number of revolutions per unit time, or the frequency (f) where f = 1/T.

How are the following formulas derived?

(Hint: use the circumference of the circle)

𝒂𝒄 =𝟒𝝅𝟐𝒓

𝑻𝟐 and 𝒂𝒄 = 𝟒𝝅

𝟐𝒓𝒇𝟐


Since 𝑣 =∆𝑑

∆𝑡 where ∆d is the circumference of a circle (C=2𝜋𝑟)

then 𝑣 =2𝜋𝑟

𝑇 where T is the period of one circle circumference.

Substituting into 𝑎𝑐=𝑣2

𝑟 =

2𝜋𝑟

𝑇2

𝑟 gives:

𝒂𝒄 =𝟒𝝅𝟐𝒓

𝑻𝟐

Since period and frequency are related as 𝑇 =1

𝑓 then:

𝒂𝒄 = 𝟒𝝅𝟐𝒓𝒇𝟐

where ac is centripetal acceleration (m/s2) r is the radius of the circular path (m) T is the period of rotation (s) f is the frequency of rotation (Hz or s-1)


Example #2: a) A salad spinner with a radius of 9.7 cm rotates clockwise

with a frequency of 12 Hz. At a given instant, a piece of

lettuce is moving in the westward direction.

Determine the magnitude and direction of the

centripetal acceleration of the lettuce in the

spinner at the moment shown.

ac = 𝟒𝝅𝟐𝒓𝒇𝟐

ac = 𝟒𝝅𝟐(𝟎. 𝟎𝟗𝟕𝒎)(𝟏𝟐𝑯𝒛)𝟐

∴ ac = 550 m/s2 [N]

b) How does the salad spinner work to remove water from the lettuce?

The water is able to pass through a screen to the outside of the spinner where it is collected, leaving the lettuce “dry”.


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 0.04 m/s2. What is its period of revolution around the sun?

ac = 𝟒𝝅𝟐𝒓

𝑻𝟐

T = 𝟒𝝅𝟐(𝟓.𝟖 𝒙 𝟏𝟎𝟏𝟎𝒎)

𝟎.𝟎𝟒 𝒎/𝒔𝟐

∴ T = 7.6 x 106 s or apprx. 88 days on Earth

Centripetal Force

According to Newton’s laws of motion, an object will accelerate only if a net force is exerted on it. Since objects moving with uniform circular motion are always accelerating, there must always be a force exerted on it in the same direction as the acceleration as shown.

The force pointing to the centre

of a circular path is called a

centripetal force (Fc). Without

this force, objects would not be able

to move in a circular path.

Centripetal Force

Using Newton’s second law and 𝑎𝑐 = 𝒗𝟐

𝒓 the formula for Fc is

derived as follows:

Substitute 𝑎𝑐 = 𝒗𝟐

𝒓 into 𝐹𝑛𝑒𝑡 = 𝑚𝑎 :

𝐹𝑛𝑒𝑡 = 𝑚𝑎𝑐

𝐹𝑛𝑒𝑡 = 𝑚𝑣2

𝑟

∴ 𝐹𝑐 = 𝐹𝑛𝑒𝑡 = 𝑚𝑣2

𝑟

Centripetal Force

SUMMARY of CENTRIPETAL FORCE

is like net force that causes centripetal acceleration (Fc =Fnet)

always choose motion towards the centre of the circle as the +ve direction

𝑭𝒄 = 𝒎𝒗𝟐

𝒓= 𝟒𝝅𝟐𝒎𝒓𝒇𝟐 =

𝟒𝝅𝟐𝒎𝒓

𝑻𝟐

where Fc is the centripetal force that acts towards the centre of circle (N)

m is the mass (kg)

Centripetal Force

A centripetal force can be supplied by any type of force. For example, gravity provides the centripetal force that keeps the Moon in a roughly circular path around Earth, friction provides a centripetal force that causes a car to move in a circular path on a flat road, and the tension in a string tied to a ball will cause the ball to move in a circular path when you twirl it around.

Centripetal Force

Example #4: An astronaut in deep space twirls a yo-yo on a string.

a) What type of force causes the yo-yo to travel

in a circle?

Tension causes Fc

b) What would happen if the string

suddenly broke?

The yo-yo would continue along a

straight line according to Newton’s

First Law of Inertia.

Centripetal Force

Example #5: A car with a mass of 2200 kg is rounding a

curve on a level road. If the radius of the curvature of the road is 52 m and the coefficient of static friction between the tires and the road is 0.70, what is the maximum speed at which the car can make the curve without skidding off the road? Since Fc = Ffs = 𝝁𝒔FN = 𝝁𝒎𝒈

and 𝑭𝒄 = 𝒎𝒗𝟐

𝒓

𝟎. 𝟕𝟎 𝟐𝟐𝟎𝟎𝒌𝒈 (𝟗. 𝟖𝟏𝒎

𝒔𝟐) =

(𝟐𝟐𝟎𝟎𝒌𝒈)𝒗𝟐

𝟓𝟐𝒎

∴ v = 19 m/s or 68 km/h

s

Centripetal Force and Banked Curves

Cars and trucks can use friction as a centripetal force. However, the amount of friction varies with road conditions and can become very small when roads are wet or icy. As well, friction causes wear and tear on tires causing them to wear out faster. For these reasons, engineers who design highways where speeds are high with large centripetal forces are required to incorporate another source of centripetal force – banked curves. Airplanes also generate a centripetal force when they bank or turn.


Example #6: A car (m = 1.1 x 103 kg) travels around a frictionless banked curve of radius 85 m. The bank is 19° to the horizontal. a) What force provides the centripetal acceleration? The horizontal component of FN acts towards the centre of the circle and results in Fc and therefore ac. Note that Ff is not needed to create Fc; only FNx creates Fc.

b) What constant speed must the car maintain to travel safely around the curve? Since 𝐹𝑥: 𝐹𝑐 = 𝑭𝑵𝒔𝒊𝒏𝜽 ; 𝒎𝒂𝒄 = 𝑭𝑵𝒔𝒊𝒏𝜽 #1

𝑭𝒚: 𝑭𝒈 = 𝑭𝑵𝒄𝒐𝒔𝜽; 𝒎𝒈 = 𝑭𝑵𝒄𝒐𝒔𝜽; 𝑭𝑵 =𝒎𝒈

𝒄𝒐𝒔𝜽 #2

Sub 2 into 1:

𝒎𝒂𝒄 =𝒎𝒈

𝒄𝒐𝒔𝜽𝒔𝒊𝒏𝜽 where mass cancels

𝒗𝟐

𝒓= 𝒈𝒕𝒂𝒏𝜽 𝒗 = 𝒓𝒈𝒕𝒂𝒏𝜽 ∴ v = 17 m/s

c) What happens if v > 17𝑚

𝑠? If v < 17

𝑚

𝑠?

Car does not maintain position; slides up or down road.

Fc


Example #7: Copy the scenario from Sample Problem 3 on pg. 122 and attempt to solve. Note that this problem incorporates friction between the tires and road to determine the maximum speed at which a car can maintain uniform circular motion.

Centrifugal Force

Sometimes when an object experiences uniform circular motion, an observer moving relative to the object may feel as though there are other forces acting on them.

Ex: On a merry-go-round you feel as though you are being pushed to the outside of the ride or while turning a corner sharply in a car or on a bike your body feels as though it leans away; from an inertial reference frame your body wants to keep moving in a straight line relative to Earth (due to inertia)

This is explained by the centrifugal force (Latin for centre-fleeing) which is a fictitious force in a non-inertial rotating frame of reference ; Newton’s Laws do not apply in an accelerating reference frame.

Centrifugal forces help to explain the perceived motion of objects in an accelerating reference frame


Read Section 3.4 on Rotating Frames of Reference and Centrifugal Forces pgs. 125-130

What challenges do long space journeys pose?

How could artificial gravity be created?

www.tcm.com/mediaroom/video/11279/2001-A-Space-Odyssey-Wide-Release-Trailer-.html

http://www.tcm.com/mediaroom/video/11279/2001-A-Space-Odyssey-Wide-Release-Trailer-.html
















Centripetal Force and Vertical Motion

Example #8: You are playing with a yo-yo of mass 225 g by swinging it vertically. The full length of the string is 1.2 m.

a) Calculate the minimum speed at which you

can swing the yo-yo while keeping it in a circular path.

Fc is caused by tension of string.

At the top of the swing if FT = 0, then v is min.

To keep the yo-yo in a circular path Fc ≥ Fg.

Since 𝑭𝒄 = 𝑭𝑻 + 𝑭𝒈 and if FT = 0

Then 𝑭𝑪 = 𝑭𝒈

𝒎𝒗𝟐

𝒓= 𝒎𝒈 where mass cancels

𝒗 = 𝟗. 𝟖𝟏𝒎/𝒔𝟐 ∗ 𝟏. 𝟐𝒎

∴ v = 3.4 m/s

Fc


Example #8 Continued: You are playing with a yo-yo (m = 225 g) by swinging it vertically. The full length of the string is 1.2 m.

b) At the speed just determined, what is the

tension in the string at the bottom of the

swing?

Since ∑F: Fc = FT - Fg

𝒎𝒗𝟐

𝒓= 𝑭𝑻 −𝒎𝒈

𝑭𝑻 =𝟎.𝟐𝟐𝟓𝒈(𝟑.𝟒𝟑𝟏𝟎𝟑𝟓)𝟐

𝟏.𝟐𝒎+ (𝟎. 𝟐𝟐𝟓𝒌𝒈 ∗ 𝟗. 𝟖𝟏

𝒎

𝒔𝟐)

𝑭𝑻 = 𝟐. 𝟐𝟎𝟕𝟐𝟓 𝑵 + 𝟐. 𝟐𝟎𝟕𝟐𝟓 𝑵

∴ FT = 4.4 N

Fc


Example #9: A roller coaster car is at the lowest point on its circular track. The radius of curvature is 22 m. The apparent weight of one of the passengers is 3.0 times her true weight (FN = 3.0Fg). Determine the speed of the roller coaster. Since 𝑭𝒄 = 𝑭𝑵 − 𝑭𝒈

𝑚𝑣2

𝑟= 3.0𝑚𝑔 − 𝑚𝑔

𝒎𝒗𝟐

𝒓= 𝟐. 𝟎𝒎𝒈

𝒗 = 𝟐𝒓𝒈 where mass cancels

𝒗 = 𝟐 ∗ 𝟐𝟐𝒎 ∗ 𝟗. 𝟖𝟏𝒎/𝒔𝟐

∴ v = 21 m/s

FN

Fg

Fc


Roller coasters have evolved over time. The circular loop that was used almost a century ago has been replaced by the clothoid loop found in modern looping coasters.

Compare these 2 designs. Refer to Section 3.5 in your text

Then…… and now…..

Roller Coaster Physics

Check out these cool links:

http://physicsbuzz.physicscentral.com/2013/04/roller-coaster-g-forces-weve-got-data.html

http://physics.gu.se/LISEBERG/eng/pendrill_loop_2013.pdf

https://www.youtube.com/watch?v=sTofF4_OPkM



















