Centripetal Acceleration 13 Examples with full solutions.

Slides:



Advertisements
Similar presentations
Horizontal Circular Motion
Advertisements

Centripetal Acceleration
Uniform Circular Motion
Force Scenario Solutions
Circular Motion and Gravitation
Chapter 10. Uniform Circular Motion
GRAVITATIONAL MOTION.
Chapter 6: Circular Motion & Other Applications of Newton’s Laws
1 Unit 6 Part 2 Circular Motion and Force. 2 Circular Motion and Centripetal Acceleration Let us take another look at our Medieval Flail. Why did the.
Uniform Circular Motion
Centripetal Acceleration and Centripetal Force
Circular Motion; Gravitation
Dynamics of Circular Motion
As the ball revolves faster, the angle increases
Motion in a Plane Chapter 8. Centripetal Acceleration Centripetal Acceleration – acceleration that points towards the center of a circle. – Also called.
Discussion Problems Thursday Feb 19 Circular motion.
CIRCULAR MOTION We will be looking at a special case of kinematics and dynamics of objects in uniform circular motion (constant speed) Cars on a circular.
Vertical Circular Motion A demo T8 T8.
5.4 highway curves 5.5 Non-uniform circular motion 5.6 Drag Velocity
Using Newton’s Laws: Friction, Circular Motion, Drag Forces
Newton’s Laws of Motion
Circular Motion.
Copyright © 2009 Pearson Education, Inc. © 2009 Pearson Education, Inc. This work is protected by United States copyright laws and is provided solely for.
1 Ch5 Circular Motion and Force. 2 Centripetal Force - Swinging Ball Any body rotating about a fixed point will experience a centripetal (center seeking)
Torque It is easier to open a door when a force is applied at the knob as opposed to a position closer to the hinges. The farther away the force, the more.
Chapter Opener. Caption: Newton’s laws are fundamental in physics
Dynamics II Motion in a Plane
AP Physics B I.E Circular Motion and Rotation. I.E.1 Uniform Circular Motion.
Example 1: A 3-kg rock swings in a circle of radius 5 m
Torque It is easier to open a door when a force is applied at the knob as opposed to a position closer to the hinges. The farther away the force, the more.
Centripetal Force and Acceleration
CIRCULAR MOTION AND OTHER APPLICATIONS OF NEWTON’S LAWS
CIRCULAR MOTION.
Circular Motion Pearson
In this chapter we will learn about the forces acting on particles when they move on a circular trajectory. Chapter 6: Circular Motion Reading assignment:
Uniform Circular Motion
CHAPTER 6 : CIRCULAR MOTION AND OTHER APPLICATIONS OF NEWTON’S LAWS
Uniform Circular Motion (UCM) The object travels in a circular path with a constant speed. Its velocity is tangent to the circle and is changing due to.
 Extension of Circular Motion & Newton’s Laws Chapter 6 Mrs. Warren Kings High School.
Uniform Circular Motion
Chapter 5 Dynamics of Uniform Circular Motion. 5.1 Uniform Circular Motion DEFINITION OF UNIFORM CIRCULAR MOTION Uniform circular motion is the motion.
Ch. 6: Circular Motion & Other Applications of Newton’s Laws
Uniform Circular Motion Centripetal forces keep these children moving in a circular path.
Copyright © 2009 Pearson Education, Inc. Chapter 5 Using Newton’s Laws: Friction, Circular Motion, Drag Forces.
Circular Motion Dynamics 8.01 W04D2. Today’s Reading Assignment: W04D2 Young and Freedman: 3.4;
Ch5 Circular Motion and Force
Circular Motion and Gravitation
Circular Motion. Period and Frequency A CD rotates in a player at a constant speed of 240 rpm. –How frequently does it complete a revolution: In minutes?
Newton’s 1 st Law of Motion. Newton’s 1 st Law Newton’s 1 st Law: An object at rest stays at rest and an object in motion stays in motion, unless acted.
Uniform Circular Motion Physics 12. Uniform Circular Motion object is moving at a constant speed but changing directions acceleration occurs due to direction.
Chapter 6.2. Uniform Circular Motion Centripetal forces keep these children moving in a circular path.
Newton’s 1 st Law of Motion. Newton’s 1 st Law Newton’s 1 st Law: An object at _______stays at _______and an object in motion stays in motion, unless.
Conceptual Physics Chapter 10
Uniform Circular Motion (UCM) The object travels in a circular path with a constant speed. Its velocity is tangent to the circle and is changing due to.
1. A car of mass 1000 kg is driving into a corner of radius 50m at a speed of 20 ms -1. The coefficient of friction between the road and the car’s tyres.
Chapter 5 Circular Motion © 2014 Pearson Education, Inc.
“What is uniform circular motion?” In uniform Circular motion a body travels at a constant speed on a circular path.
Centripetal Acceleration
When you ride the Spindletop, you move in a circle at a constant speed
Uniform Circular Motion
Uniform Circular Motion
Vertical Circular Motion
Aim: How do we explain centripetal motion?
More Centripetal Force Problems
Centripetal forces keep these children moving in a circular path.
Vertical Circular Motion
5-2 Uniform Circular Motion—Kinematics
Aim: How do we explain centripetal motion?
Pendulum A string is attached to mass m and swung in a horizontal circle of radius r. There is tension T in the string. The angle θ is measured from the.
Presentation transcript:

Centripetal Acceleration 13 Examples with full solutions

Example 1 A 1500 kg car is moving on a flat road and negotiates a curve whose radius is 35m. If the coefficient of static friction between the tires and the road is 0.5, determine the maximum speed the car can have in order to successfully make the turn. 35m

Example 1 – Step 1 (Free Body Diagram) +y +x This static friction is the only horizontal force keeping the car moving toward the centre of the arc (else the car will drive off the road). Acceleration direction

Example 1 - Step 2 (Sum of Vector Components ) +y +x Vertical ComponentsHorizontal Components We have an acceleration in x-direction Static Friction From Vertical Component

Example 1 - Step 3 (Insert values) +y +x

Example 2 A car is travelling at 25m/s around a level curve of radius 120m. What is the minimum value of the coefficient of static friction between the tires and the road to prevent the car from skidding? 120 m

Example 2 – Step 1 (Free Body Diagram) +y +x This static friction is the only horizontal force keeping the car moving toward the centre of the arc (else the car will drive off the road). Acceleration direction

Example 2 - Step 2 (Sum of Vector Components ) +y +x Vertical ComponentsHorizontal Components We have an acceleration in x-direction Static Friction From Vertical Component

Example 2 - Step 3 (Insert values) +y +x We require the minimum value

Example 3 An engineer has design a banked corner with a radius of 200m and an angle of What should the maximum speed be so that any vehicle can manage the corner even if there is no friction? m

Example 3 – Step 1 (Free Body Diagram) First the car Now for gravity The normal to the road Components of Normal force along axis (we ensured one axis was along acceleration direction +y +x Acceleration direction Notice that we have no static friction force in this example (question did not require one)

Example 3 - Step 2 (Sum of Vector Components ) +y +x Vertical Components Horizontal Components We have an acceleration in the x-direction From Vertical Component

Example 3 - Step 3 (Insert values) +y +x

Example 4 An engineer has design a banked corner with a radius of 230m and the bank must handle speeds of 88 km/h. What bank angle should the engineer design to handle the road if it completely ices up? 230 m ?

Example 4 – Step 1 (Free Body Diagram) First the car Now for gravity The normal to the road Components of Normal force along axis (we ensured one axis was along acceleration direction +y +x Acceleration direction Notice that we have no static friction force in this example (question did not require one)

Example 4 - Step 2 (Sum of Vector Components ) +y +x Vertical Components Horizontal Components We have an acceleration in the x-direction From Vertical Component

+y +x Example 4 - Step 3 (Insert values) Don’t forget to place in metres per second

Example 5 A 2kg ball is rotated in a vertical direction. The ball is attached to a light string of length 3m and the ball is kept moving at a constant speed of 12 m/s. Determine the tension is the string at the highest and lowest points.

Example 5 – Step 1 (Free Body Diagram) Top Bottom When the ball is at the top of the curve, the string is “pulling” down. When the ball is at the bottom of the curve, the string is “pulling” up. In both cases, gravity is pulling down Note: any vertical motion problems that do not include a solid attachment to the centre, do not maintain a constant speed, v and thus (except at top and bottom) have an acceleration that does not point toward the centre. (it is better to use energy conservation techniques)

Example 5 - Step 2 (Sum of Vector Components ) Top Bottom +y +x Note: acceleration is down (-)

Example 5 - Step 3 (Insert values) +y +x Top Bottom

Example 6 A conical pendulum consists of a mass (the pendulum bob) that travels in a circle on the end of a string tracing out a cone. If the mass of the bob is 1.7 kg, and the length of the string is 1.25 m, and the angle the string makes with the vertical is 25 o. Determine: a) the speed of the bob b) the frequency of the bob

Example 6 – Step 1 (Free Body Diagram) Let’s decompose our Tension Force into vertical and horizontal components +y +x It’s easier to make the x axis positive to the left

+y +x Example 6 - Step 2 (Sum of Vector Components ) Horizontal Vertical

Example 6 - Step 3 (Insert values for velocity) +y +x The speed of the bob is about 1.55 m/s

Example 6 - Step 3 (Insert values for frequency) +y +x The frequency of the bob is about 0.468Hz

Example 7 A swing at an amusement park consists of a vertical central shaft with a number of horizontal arms. Each arm supports a seat suspended from a cable 5.00m long. The upper end of the cable is attached to the arm 3.00 m from the central shaft. Determine the time for one revolution of the swing if the cable makes an angle of 30 0 with the vertical

Example 7 – Step 1 (Free Body Diagram) +y +x

Example 7 - Step 2 (Sum of Vector Components ) +y +x Horizontal Vertical

Example 7 - Step 2 (Sum of Vector Components ) +y +x The period is 6.19s

Example 8 A toy car with a mass of 1.60 kg moves at a constant speed of 12.0 m/s in a vertical circle inside a metal cylinder that has a radius of 5.00 m. What is the magnitude of the normal force exerted by the walls of the cylinder at A the bottom of the circle and at B the top of the circle

Example 8 – Step 1 (Free Body Diagram) Top Bottom When the car is at the top of the curve, the normal force is “pushing” down. When the ball is at the bottom of the curve, the normal force is “pushing” up. In both cases, gravity is pulling down Note: any vertical motion problems that do not include a solid attachment to the centre, do not maintain a constant speed, v and thus (except at top and bottom) have an acceleration that does not point toward the centre. (it is better to use energy conservation techniques)

Example 8 - Step 2 (Sum of Vector Components ) Top Bottom +y +x Note: acceleration is down (-)

Example 8 - Step 3 (Insert values) +y +x Top Bottom

Example 9 A 0.20g fly sits 12cm from the centre of a phonograph record revolving at rpm. a) What is the magnitude of the centripetal force on the fly? b) What is the minimum static friction between the fly and the record to prevent the fly from sliding off?

Example 8 – Step 1 (Free Body Diagram)

Example 9 - Step 2 (Sum of Vector Components ) a. Convert to correct units

Example 9 - Step 2 (Sum of Vector Components ) b.

Example 10 A 4.00 kg mass is attached to a vertical rod by the means of two 1.25 m strings which are 2.00 m apart. The mass rotates about the vertical shaft producing a tension of 80.0 N in the top string. a)What is the tension on the lower string? b)How many revolutions per minute does the system make?

Example 10 – Step 1 (Free Body Diagram) +y +x

Example 10 - Step 2 (Sum of Vector Components) Horizontal Vertical +y +

Example 10 - Step 2 (Sum of Vector Components ) +y + a)

Example 10 - Step 2 (Sum of Vector Components ) +y + b)

Example 11 The moon orbits the Earth in an approximately circular path of radius 3.8 x 10 8 m. It takes about 27 days to complete one orbit. What is the mass of the Earth as obtained by this data?

Example 11 – Step 1 (Free Body Diagram)

Example 11 - Step 2 (Sum of Vector Components) The mass of the Earth is about 6.0x10 24 kg Horizontal

Example 12 Cassiopia takes a ride on child’s Ferris Wheel. This ride has no retaining bar, so that she only sides on the seat as the ride moves. a)Determine the Normal Force she would experience from the bottom of the seat when she is at the lowest point on the ride. b)Determine the Normal Force she would experience from the bottom of the seat when she is at the highest point on the ride. c)Determine the Net Force she would experience from the bottom of the seat when she is at the mid-point on the ride with her height equal to the axis.

Cassiopia takes a ride on child’s Ferris Wheel. This ride has no retaining bar, so that she only sides on the seat as the ride moves. a)Determine the Normal Force she would experience from the bottom of the seat when she is at the lowest point on the ride. With this ride, gravity always points down, the normal (seat) force always points up, and the centripetal acceleration is always toward the centre.

Cassiopia takes a ride on child’s Ferris Wheel. This ride has no retaining bar, so that she only sides on the seat as the ride moves. Determine the Normal Force she would experience from the bottom of the seat when she is at the highest point on the ride. With this ride, gravity always points down, the normal (seat) force always points up, and the centripetal acceleration is always toward the centre.

Example 12 Cassiopia takes a ride on child’s Ferris Wheel. This ride has no retaining bar, so that she only sides on the seat as the ride moves. Determine the Net Force (force of seat) she would experience from the bottom of the seat when she is at the mid-point on the ride with her height equal to the axis.

Example 13 (Hard Question) An engineer has design a banked corner with a radius of R and an angle of β. What is the equation that determines the velocity of the car given that the coefficient of friction is µ ?

Example 13 – Step 1 (Free Body Diagram) First the car Now for gravity The normal to the road Components of Normal force along axis (we ensured one axis was along acceleration direction +y +x Acceleration direction Friction We have friction going down by assuming car wants to slide up. This will provide an equations for the maximum velocity

Vertical Components Horizontal Components Example 13 – Step 2 (Components) +y +x+x From Vertical

Sub into Horizontal Example 13 – Step 2 (Components) +y +x+x From Vertical Solve for v π Minimum velocity (slides down)

Flash