chapter 5, lect. 10 additional applications of newton’s laws

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 .