1. Chapter 5, Lect. 10 Additional Applications of Newton’s Laws
Today: circular motion, center of mass
When and where
TODAY: 5:45-7:00 pm
Rooms: See course webpage. Be sure report to your TA’s room
Your TA will give a review during the discussion session next week.
Format
Closed book, 20 multiple-choices questions (consult with practice exam)
One-page formula sheet allowed, must be self prepared, no photo copying/download-printing of solutions, lecture slides, etc.
Bring a calculator (but no computer). Only basic calculation functionality
can be used. Bring a 2B pencil for Scantron.
Fill in your ID and section # !
Special requests:
One alternative exam all set:
3:30pm – 4:45pm, Thurs Feb.17, room Chamberlin (!).
About Midterm Exam 1
4/24/2017
Phys 201, Spring 2011

2. impossible to tell from the given information.
The figure shows a top view of a ball on the end of a string traveling counterclockwise in a circular path. The speed of the ball is constant. If the string should break at the instant shown, the path that the ball would follow is 1 2 3 4
impossible to tell from the given information.
4/24/2017
Phys 201, Spring 2011

3. Acceleration on a curved path
Instead of considering
a = ax i + ay j + az k (time-independent)
Decomposed into:
a = at + ac
Tangential acceleration: at = dv/dt
The magnitude change of v.
Centripetal acceleration: ac
The direction change of v.
Phys 201, Spring 2011
4/24/2017

4. Centripetal acceleration is the acceleration perpendicular to the velocity that occurs when a particle is moving on a curved path. 
Centripetal force associated with centripetal acceleration, directed towards the center of the circle:
4/24/2017
Phys 201, Spring 2011

5. Uniform Circular Motion
If object is moving with constant speed on the circle, v = const. r = const. ac = v2/r = const.
Motion in a Horizontal Circle
The centripetal force is supplied by the tension
4/24/2017
Phys 201, Spring 2011

6. Example:
An object of mass m is suspended from a point in the ceiling on a string of length L. The object revolves with constant speed v in a horizontal circle of radius r. (The string makes an angle θ with the vertical).
The speed v is given by the expression: θ L y
4/24/2017
Phys 201, Spring 2011

7. Horizontal (Flat) Curve
The force of static friction supplies the centripetal force
The maximum speed at which the car can negotiate the curve is
Note, this does not depend on the mass of the car
4/24/2017
Phys 201, Spring 2011

8. (2/3)ac (4/3)ac (2/9)ac (9/2)ac (3/2)ac
A car going around a curve of radius R at a speed V experiences a centripetal acceleration ac. What is its acceleration if it goes around a curve of radius 3R at a speed of 2V?
(2/3)ac
(4/3)ac
(2/9)ac
(9/2)ac
(3/2)ac
4/24/2017
Phys 201, Spring 2011

9. Banked Curve
These are designed to be navigable when there is no friction
There is a component of the normal force that supplies the centripetal force (even μ=0!)
nx =
ny =
4/24/2017
Phys 201, Spring 2011

10. Non-Uniform Circular Motion
The acceleration and force have tangential components
Fr produces the centripetal acceleration (change v in directions)
Ft produces the tangential acceleration (v change in magnitude)
ΣF = ΣFr + ΣFt
4/24/2017
Phys 201, Spring 2011

11. Vertical Circle With Non-Uniform Speed
The gravitational force exerts a tangential force on the object
Look at the components of Fg
The tension at any point:
along ac direction:
ac = T - gravity
4/24/2017
Phys 201, Spring 2011

12. Top and Bottom of Circle
The tension at the bottom is a maximum: cosθ = +1
The tension at the top is a minimum: cosθ = -1
If Ttop = 0, gravity does it:
4/24/2017
Phys 201, Spring 2011

13. The Center of Mass Definition of center of mass: Where
For a continuous object (e.g., a solid sphere)
4/24/2017
Phys 201, Spring 2011

14. CM position for a semicircular hoop
where M = λπR,
CM position can be outside the body.
4/24/2017
Phys 201, Spring 2011

15. Center of Mass (2) Newton’s Laws for a collection of objects:
The acceleration of the center of mass is determined entirely by the external net force
on the objects.
4/24/2017
Phys 201, Spring 2011

16. Changing Places in a Rowboat : Fnet Ext = 0  Xcm fixed
(initial condition: 0)
mP XP + mD XD + mb Xb = 0
mP X’P + mD X’D + mb X’b = 0
mP ΔXP + mD ΔXD + mb ΔXb = 0
with ΔXP = -ΔXD = L
Thus, (mP – mD) L = -mb ΔXb
ΔXb = L (mD – mP)/mb .
4/24/2017
Phys 201, Spring 2011