announcements :

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Announcements : - Midterm exam coming up Wednesday Feb. 16, (two evening times, 5-6 pm or 6-7 pm). - Material: Chapter 1-5. - Key equations will be provided on the last page of the exam. - Bring a calculator to the test (non- programmable). - I will put up a practice exam on our class web page ( http://www.wfu.edu/~gutholdm/Physics113/phy113.html )

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Announcements : Midterm exam coming up Wednesday Feb. 16 , (two evening times, 5-6 pm or 6-7 pm). Material: Chapter 1-5 . Key equations will be provided on the last page of the exam. Bring a calculator to the test ( non-programmable ). - PowerPoint PPT Presentation

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Page 1: Announcements :

Announcements:- Midterm exam coming up Wednesday Feb. 16, (two evening

times, 5-6 pm or 6-7 pm).- Material: Chapter 1-5. - Key equations will be provided on the last page of the exam. - Bring a calculator to the test (non-programmable).- I will put up a practice exam on our class web page (

http://www.wfu.edu/~gutholdm/Physics113/phy113.html)

Page 2: Announcements :

• 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: Chapter 4.4, 4.5, 6.1, 6.2 (skip 6.3, 6.4, 6.5)

Homework : Ch. 4: OQ1, OQ7, 27, 30, 32

Ch.6: 1, 6, 8, 9, 57

Due dates: Tu/Th section: Monday, Feb. 21

MWF section: Monday, Feb. 21

Remember: Homework 5 is due Monday, Feb. 14 (section A & B)

Page 3: Announcements :

Quick quizzName all objects that experience a net accelerationA. Space ship at constant velocityB. A braking carC. Particle going straight at constant speedD. Hammer of a hammer thrower (see picture)E. B & D

Page 4: Announcements :

Uniform Circular Motion

Motion in a circular path at constant speed.• Velocity is changing, thus there is an acceleration!!• Velocity is tangent to the path of the object• Acceleration is perpendicular to velocity• Centripetal acceleration is towards the center of the circle • Magnitude of acceleration is • r is radius of circle r

vac

2

Page 5: Announcements :

Thus far we have applied Newton’s law, F = m*a to linear motion.

Now we’ll apply it to rotational motion

Page 6: Announcements :

Particle moving with uniform speed v in a circular path with radius r has an acceleration ar:

rvar

2

(Derivation: see Chapter 4.4)

- The acceleration points towards the center of the circle!

- Centripetal acceleration

Page 7: Announcements :

Newton’s law along the radial direction (along r):

rvmamF c

2

Page 8: Announcements :

Uniform Circular motion:• The velocity of the particle is along the __________• The centripetal acceleration is towards the __________• The centripetal force acting on the particle is towards the

________

The magnitude of the centripetal acceleration is: a =______________

Newton’s law: The force on the particle is (centripetal force)

F= m·a = ______________

• Centripetal force causes a change in the ________________ but no change in ________________.

Page 9: Announcements :

A particle is moving in a circular path.

If the force on the particle would suddenly vanish (string cut) in which direction would the ball fly off?

Page 10: Announcements :

Black board example 6.1

Jeff Gordon leads his race and must drive into a curve at top speed to win it all.

The radius of the curve is 1000.0 m and the coefficient of static friction between his tires and the pavement is 0.500. Find the maximum speed he can have and still make the turn.

Page 11: Announcements :

An amusement park ride consists of a rotating circular platform 8.00 m in diameter from which 10.0 kg seats are suspended from 2.5 m chains. When the system rotates, the chains make an angle of 28° with the vertical.

Black board example 6.2

a) Draw a free body diagram of a 40.0 kg child riding in a seat and find the tension in the chain.

b) What is the speed of each seat?

Page 12: Announcements :

Non-uniform circular motion, tangential and radial acceleration

Now: consider particles which move on a circular path with changing speed (speeding up or slowing down). Total acceleration has a tangential and radial component:

• Tangential component causes a change in speed• Radial component causes a change in the direction of the velocity

rva

dtvd

a

aaa

r

t

tr

2

Page 13: Announcements :

Non-uniform circular motion, tangential and radial acceleration

Previous slide: acceleration has a radial and tangential component

Force has a corresponding radial and tangential component.

The bead speeds up with constant tangential acceleration, as it moves right.

Draw the vectors representing the force on the bead at points A, B, C.