Download presentation

1
**Chapter 5: Uniform Circular Motion**

Chapter Goal: To learn how to solve problems about uniform circular motion. Uniform circular motion: the motion of an object traveling at a constant speed on a circular path. The speed may be constant, but the velocity is not.

2
**Uniform Circular Motion**

The velocity vector shows the direction of motion at any point on the circle. For uniform circular motion the velocity vectors at each point are tangent to the circle and are the same length Period (T) – the time it takes to make one revolution around the circle. T is in seconds, v is in m/s.

3
Example Problem Your roommate is working on his bicycle. The bicycle wheel has a radius of 0.3 meters. As he spins the bicycle wheel, you notice that a stone is stuck in the tread. It goes by 3 times every second. a. What is the period of the stone? What is the speed of the stone? Hint: the stone makes 3 revolutions per second (rps). The period is how many seconds it takes to make one revolution.

4
**Example Problem - Answer**

Your roommate is working on his bicycle. The bicycle wheel has a radius of 0.3 meters. As he spins the bicycle wheel, you notice that a stone is stuck in the tread. It goes by 3 times every second. a. What is the period of the stone? T = 1/3 rps = .333s. Note that period, T is the reciprocal of rps. What is the speed of the stone? v = 2πr/T = 2π (.3m)/(.333s) = 5.7m/s

5
**Determining the direction of acceleration for uniform circular motion using vector subtraction**

Draw a velocity vector on top of each even dot, tangent to the circle. Starting with 0, take each even number and do a graphical vector subtraction to find the direction of ∆v and therefore Locate the ∆v vector at the odd number midway between the 2 velocity vectors.

6
**Acceleration of Uniform Circular motion**

If some net force results in uniform circular motion, the resulting acceleration always points towards the center of the circle. This is called centripetal (“center-seeking”) acceleration. Is centripetal acceleration equal to ∆v/ ∆t? Yes, but if the magnitude of v remains constant, how do we evaluate ∆v ? Using trig, it can be shown that ac = ∆v/ ∆t = v2/r.

7
Rank in order, from largest to smallest, the centripetal accelerations (ac)a to (ac)e of particles a to e. Answer: B

8
**Determining the direction of acceleration for uniform circular motion using Newton’s 2nd Law**

ΣF = ma tells us the net force vector and the acceleration vector are in the same direction (since mass is a scalar). Take a ball with a string attached and roll it around on the table in a perfect (!) circle at constant speed. How do we draw a free body diagram of something that is constantly changing direction? The radius becomes the x axis (actually the radial axis). This means “toward the center of the circle” is always considered positive. The y axis is perpendicular to the plane of motion.

9
**FBD of a ball on a string in uniform circular motion**

Draw a freebody diagram of the ball on a string in uniform circular motion. In which direction is Fnet? In which direction is a? How does this change as the ball goes around the circle?

10
**FBD of a ball on a string in uniform circular motion**

y ΣFy = may = 0 (ball is not hopping up or crashing down) ΣFx = max = T Since m is a scalar, T and a must point in the same direction, toward the center of the circle. n T x Fg a

11
**FBD of a ball on a track in uniform circular motion**

Draw a freebody diagram of the ball on a track in uniform circular motion. In which direction is Fnet? In which direction is a? How does this change as the ball goes around the circle?

12
**FBD of a ball on a track in uniform circular motion**

y ΣFy = may = 0 (ball is not hopping up or crashing down) ΣFx = max = n2 Since m is a scalar, n2 and a must point in the same direction, toward the center of the circle. n1 n2 x Fg a

13
**Car turning a “circular” corner at constant speed**

Is there a net force on the car as it negotiates the turn? Where did it come from?

14
**Car turning a “circular” corner at constant speed**

When the driver turns the wheel the tires turn. To continue along a straight line, the car must overcome static friction and slide. If the static friction force is less than the maximum, the tire cannot slide and so has no choice but to roll in the direction of the turn. fs-max = μs |n| skid…. blue arrow represents direction car would slide in the absence of friction

15
**Car turning a “circular” corner at constant speed – example problem**

What is the maximum speed at which a 1500 kg car can make a turn around a curve of radius 50 m on a level road without sliding out of the turn (skidding)? Recall toward the center of the circle is positive, even if it is to the left in this example

16
**Car turning a “circular” corner at constant speed – example problem**

What is the maximum speed at which a 1500 kg car can make a turn around a curve of radius 50 m on a level road without sliding out of the turn (skidding)? 1. draw a free body diagram with appropriate axes and lists knowns and “finds”

17
**Car turning a “circular” corner at constant speed – example problem**

Newton’s 2nd Law in component form. Start with y (z) axis since there is no acceleration: ΣFy = may = 0 n – FG = 0, therefore n = mg

18
**Car turning a “circular” corner at constant speed – example problem**

Newton’s 2nd Law in component form. Now do forces in the x (r) direction: ΣFx = mac = mv2/r. The only force in the x direction is the frictional force, therefore: fs = mv2/r vmax occurs at fs-max. Therefore: μs |n| = mv2/r or μs mg = mv2/r

19
**Car turning a “circular” corner at constant speed – example problem**

Note the mass drops out, as all terms in the equation include mass. Solve for vmax : v = (us rg)1/2 = 22 m/s, about 45 mph Slow down!

20
**Example – comparing speeds**

Car A uses tires for which the coefficient of static friction is 1.1 on a particular unbanked curve. The maximum speed at which the car can negotiate this curve is 25 m/s. Car B uses tires for which the coefficient of static friction is 0.95 on the same curve. What is the maximum speed at which car B can negotiate the curve? Use Newton’s Law to find a in terms of µs

21
**Example – comparing speeds**

Car A uses tires for which the coefficient of static friction is 1.1 on a particular unbanked curve. The maximum speed at which the car can negotiate this curve is 25 m/s. Car B uses tires for which the coefficient of static friction is 0.95 on the same curve. What is the maximum speed at which car B can negotiate the curve? Known vA = 25m/s µA = 1.1 µB = 0.95 Find vB

22
**Example – comparing speeds**

Car A uses tires for which the coefficient of static friction is 1.1 on a particular unbanked curve. The maximum speed at which the car can negotiate this curve is 25 m/s. Car B uses tires for which the coefficient of static friction is 0.95 on the same curve. What is the maximum speed at which car B can negotiate the curve? Since the radius is the same in both cases, and g is a constant, use ratio reasoning to calculate vB

23
**Example – comparing speeds**

Car A uses tires for which the coefficient of static friction is 1.1 on a particular unbanked curve. The maximum speed at which the car can negotiate this curve is 25 m/s. Car B uses tires for which the coefficient of static friction is 0.95 on the same curve. What is the maximum speed at which car B can negotiate the curve? Known vA = 25m/s µA = 1.1 µB = 0.95 Find vB

24
**Example – comparing speeds**

Car A uses tires for which the coefficient of static friction is 1.1 on a particular unbanked curve. The maximum speed at which the car can negotiate this curve is 25 m/s. Car B uses tires for which the coefficient of static friction is 0.95 on the same curve. What is the maximum speed at which car B can negotiate the curve? Known vA = 25m/s µA = 1.1 µB = 0.95 Find vB

25
**Example – comparing speeds**

Car A uses tires for which the coefficient of static friction is 1.1 on a particular unbanked curve. The maximum speed at which the car can negotiate this curve is 25 m/s. Car B uses tires for which the coefficient of static friction is 0.95 on the same curve. What is the maximum speed at which car B can negotiate the curve? vB = 23 m/s

26
**Example – think fuzzy dice**

A block is hung by a string from the inside roof of a van. When the van goes straight ahead at a speed of 28 m/s, the block hangs vertically down. But when the van maintains this same speed around an unbanked curve (radius = 150 m), the block swings toward the outside of the curve. Then the string makes an angle θ with the vertical. Find θ. θ

27
Example problem A penny is placed 15 cm from the center of a turntable. The table has period of 1.5 s. What is the mimimum coefficient of static friction so that the penny will rotate with the turntable and not slide off?

28
**Example problem - answer**

A penny is placed 15 cm from the center of a turntable. The table has period of 1.5 s. What is the mimimum coefficient of static friction so that the penny will rotate with the turntable and not slide off? Known r = 15 cm T = 1.5 s Find µsmax

29
EOC # 23 A ride at a carnival consists of chairs that are swung in a circle by 15.0-m cables attached to a vertical rotating pole, as shown in the drawing. The total mass of a chair and occupant is 179 kg. a) Draw a free body diagram for this situation. Note that angle is from vertical, but the Indian Princess can cope. Oh, and what is the radius in this problem? b) Determine the Tension in cable c) Determine speed of chair 29

30
EOC # 23 A ride at a carnival consists of chairs that are swung in a circle by 15.0-m cables attached to a vertical rotating pole, as shown in the drawing. The total mass of a chair and occupant is 179 kg. ΣFy = 0, no acceleration in y direction. ΣFx = mac where ac = mv2/r 60° Known L = 15.0 m, r = L sin θ m = 179 kg θ = 60° (from vertical) Find v, T 30

31
A ride at a carnival consists of chairs that are swung in a circle by 15.0-m cables attached to a vertical rotating pole, as shown in the drawing. The total mass of a chair and occupant is 179 kg. 3508 N 14.8 m/s 60° 31

32
**Friction in the back seat**

A woman is sitting in the middle of the back seat of the car. The car rounds a 60-m radius curve while moving at a constant speed of 20 m/s. What is the minumum coefficient of static friction between her pants and the car seat so that she doesn’t slide across the seat? Draw a fbd and write down Newton’s second law in component form.

33
Stone in the tire tread A stone has a mass of kg and is wedged into the tread of an automobile tire, as the drawing shows. The coefficient of static friction between the stone and each side of the tread channel is When the tire surface is rotating at a maximum speed of 13 m/s, the stone flies out of the tread. The magnitude FN of the normal force that each side of the tread channel exerts on the stone is 1.8 N. Assume that only static friction supplies the centripetal force, and determine the radius r of the tire.

34
**Friction in the back seat**

A woman is sitting in the middle of the back seat of the car. The car rounds a 60-m radius curve while moving at a constant speed of 20 m/s. What is the minimum coefficient of static friction between her pants and the car seat so that she doesn’t slide across the seat? Draw a fbd and write down Newton’s second law in component form. r=60m, v = 20m/s Find us ΣFy = may = 0 ΣFx = mac = mv2/r. She refused to tell me her mass, so you have to solve without it…

35
**Friction in the back seat**

A woman is sitting in the middle of the back seat of the car. The car rounds a 60-m radius curve while moving at a constant speed of 20 m/s. What is the minimum coefficient of static friction between her pants and the car seat so that she doesn’t slide across the seat? Draw a fbd and write down Newton’s second law in component form. r=60m, v = 20m/s Find us ΣFy = may = 0 ΣFx = mac = mv2/r. μs = 0.68

Similar presentations

OK

Circular Motion Part 2 By: Heather Britton. Circular Motion Part 2 According to Newton’s 2nd Law, an accelerating body must have a force acting on it.

Circular Motion Part 2 By: Heather Britton. Circular Motion Part 2 According to Newton’s 2nd Law, an accelerating body must have a force acting on it.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on charge coupled device for side Ppt on non agricultural activities definition Ppt on word association test interpretation Ppt on precautions of tsunami Ppt on phonetic transcription Ppt on obesity management protocols Download ppt on coordinate geometry for class 9th notes Ppt on polynomials free download Ppt on eid festival in pakistan Ppt on trade related intellectual property rights