Chapter 5, Lect. 10Additional Applications of Newton’s Laws
Today: circular motion, center of mass
04/24/23 Phys 201, Spring 2011
• 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 5280 Chamberlin (!).
About Midterm Exam 1
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
04/24/23 Phys 201, Spring 2011
Phys 201, Spring 2011
Acceleration on a curved path
Decomposed into: a = at + ac
Tangential acceleration: at = dv/dtThe magnitude change of v.
Centripetal acceleration: ac
The direction change of v.
04/24/23
Instead of considering aa = ax ii + ay jj + az k k (time-independent)
• 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:
04/24/23 Phys 201, Spring 2011
Uniform Circular MotionIf object is moving with constant speed on the circle, v = const. r = const. ac = v2/r = const.
Motion in a Horizontal CircleThe centripetal force is supplied by the tension
T =mv2
r⇒
04/24/23 Phys 201, Spring 2011
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:
y
L
€
r F net =m
r a
x : Tsinθ =m v2
ry: Tcosθ−mg =0
⇒ tanθ =v2
rg⇒ v= rgtanθ
θ
04/24/23 Phys 201, Spring 2011
Horizontal (Flat) Curve The force of static friction
supplies the centripetal force
The maximum speed at which the car can negotiate the curve is
fs =mv2
rfsmax =m
vm ax2
r=msmg
Note, this does not depend on the mass of the car04/24/23 Phys 201, Spring 2011
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?
04/24/23 Phys 201, Spring 2011
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!)
ncosθ =mg
nsinθ =mv2
rny =
nx =
04/24/23 Phys 201, Spring 2011
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
04/24/23 Phys 201, Spring 2011
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
04/24/23 Phys 201, Spring 2011
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:
04/24/23 Phys 201, Spring 2011
The Center of Mass
• Definition of center of mass:
Where
For a continuous object (e.g., a solid sphere)
04/24/23 Phys 201, Spring 2011
04/24/23 Phys 201, Spring 2011
CM position for a semicircular hoop
where M = λπR,
CM position can be outside the body.
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.
04/24/23 Phys 201, Spring 2011
04/24/23 Phys 201, Spring 2011
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 .