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PHY 113 C General Physics I 11 AM – 12:15 PM TR Olin 101 Plan for Lecture 17: Review of Chapters 9-13, 15-16 Comment on exam and advice for preparation Review Example problems. Webassign questions – Assignment #15

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11 AM – 12:15 PM TR Olin 101

Plan for Lecture 17:

Review of Chapters 9-13, 15-16

Comment on exam and advice for preparation

Review

Example problems

PHY 113 C Fall 2013 -- Lecture 17

Webassign questions – Assignment #15

Consider the sinusoidal wave of the figure below with the wave function y = 0.150 cos(15.7x − 50.3t)

where x and y are in meters and t is in seconds. At a certain instant, let point A be at the origin and point B be the closest point to A along the x axis where the wave is 43.0° out of phase with A. What is the coordinate of B?

PHY 113 C Fall 2013 -- Lecture 17

Webassign questions – Assignment #15

A transverse wave on a string is described by the following wave function. y = 0.115 sin ((π/9)x+ 5πt)

where x and y are in meters and t is in seconds.

Determine the transverse speed at t = 0.150 s for an element of the string located at x = 1.50 m.

(b) Determine the transverse acceleration at t = 0.150 s for an element of the string located at x = 1.50 m.

PHY 113 C Fall 2013 -- Lecture 17

Webassign questions – Assignment #15

A sinusoidal wave in a rope is described by the wave function

y= 0.20 sin (0.69πx + 20πt)

where x and y are in meters and t is in seconds. The rope has a linear mass density of 0.230 kg/m. The tension in the rope is provided by an arrangement like the one illustrated in the figure below. What is the mass of the suspended object?

T

mg

PHY 113 C Fall 2013 -- Lecture 17

PHY 113 C Fall 2013 -- Lecture 17

• iclicker question

• What is the purpose of exams?

• Pure pain and suffering for all involved.

• To measure what has been learned.

• To help students learn the material.

• Other.

PHY 113 C Fall 2013 -- Lecture 17

• Topics covered

• Linear momentum

• Rotational motion and angular momentum

• Gravitational force and circular orbits

• Static equilibrium

• Simple harmonic motion

• Wave motion

PHY 113 C Fall 2013 -- Lecture 17

PHY 113 C Fall 2013 -- Lecture 17

iclicker question:

Have you looked at last year’s exams?

A. Yes B. No

PHY 113 C Fall 2013 -- Lecture 17

• Linear momentum

• What is it?

• When is it “conserved”?

• Conservation of momentum in analysis of collisions

• Notion of center of mass

PHY 113 C Fall 2013 -- Lecture 17

Physics of composite systems

PHY 113 C Fall 2013 -- Lecture 17

Example – completely inelastic collision; balls moving in one dimension on a frictionless surface

PHY 113 C Fall 2013 -- Lecture 17

Examples of two-dimensional collision; balls moving on a frictionless surface

PHY 113 C Fall 2013 -- Lecture 17

The notion of the center of mass and the physics of composite systems

PHY 113 C Fall 2013 -- Lecture 17

Finding the center of mass composite systems

PHY 113 C Fall 2013 -- Lecture 17

q

PHY 113 C Fall 2013 -- Lecture 17

Review of rotational energy associated with a rigid body composite systems

PHY 113 C Fall 2013 -- Lecture 17

Moment of inertia: composite systems

PHY 113 C Fall 2013 -- Lecture 17

CM composite systems

CM

PHY 113 C Fall 2013 -- Lecture 17

iclicker composite systems exercise:

Three round balls, each having a mass M and radius R, start from rest at the top of the incline. After they are released, they roll without slipping down the incline. Which ball will reach the bottom first?

C

B

A

PHY 113 C Fall 2013 -- Lecture 17

q composite systems

How can you make objects rotate?

Define torque:

t = r x F

t = rF sin q

r

F sin q

q

F

PHY 113 C Fall 2013 -- Lecture 17

Example form composite systemsWebassign #11

• iclicker exercise

• When the pivot point is O, which torque is zero?

• A. t1?

• B. t2?

• C. t3?

t3

X

t2

t1

PHY 113 C Fall 2013 -- Lecture 17

Vector cross product; right hand rule composite systems

PHY 113 C Fall 2013 -- Lecture 17

From Newton’s second law – continued – conservation of angular momentum:

PHY 113 C Fall 2013 -- Lecture 17

Example of conservation of angular momentum angular momentum:

PHY 113 C Fall 2013 -- Lecture 17

Summary – conservation laws we have studied so far angular momentum:

PHY 113 C Fall 2013 -- Lecture 17

PHY 113 C Fall 2013 -- Lecture 17

Universal law of gravitation angular momentum:

 Newton (with help from Galileo, Kepler, etc.) 1687

PHY 113 C Fall 2013 -- Lecture 17

Gravitational force of the Earth angular momentum:

RE

m

Note: Earth’s gravity acts as a point mass located at the Earth’s center.

PHY 113 C Fall 2013 -- Lecture 17

R angular momentum:EM

F

Stable circular orbit of two gravitationally attracted objects (such as the moon and the Earth)

v

a

PHY 113 C Fall 2013 -- Lecture 17

m angular momentum:2

R2

R1

m1

Circular orbital motion about center of mass

v2

CM

v1

PHY 113 C Fall 2013 -- Lecture 17

m angular momentum:2

R2

R1

m1

v2

L1=m1v1R1

L2=m2v2R2

L = L1 + L2

v1

Note: More generally, stable orbits can be elliptical.

PHY 113 C Fall 2013 -- Lecture 17

Gravitational potential energy angular momentum:

Example:

PHY 113 C Fall 2013 -- Lecture 17

Analysis of static equilibrium angular momentum:

Meanwhile – back on the surface of the Earth:

Conditions for stable equilibrium

PHY 113 C Fall 2013 -- Lecture 17

X angular momentum:

**

T

Mg

mg

PHY 113 C Fall 2013 -- Lecture 17

Some practice problems angular momentum:

PHY 113 C Fall 2013 -- Lecture 17

From angular momentum:webassign:

A 100-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.800 rev/s in 2.00 s? (State the magnitude of the force.)

view from top:

F

R

PHY 113 C Fall 2013 -- Lecture 17

From angular momentum:webassign:

A 10.3-kg monkey climbs a uniform ladder with weight w = 1.24  102 N and length L = 3.35 m as shown in the figure below. The ladder rests against the wall and makes an angle of θ = 60.0° with the ground. The upper and lower ends of the ladder rest on frictionless surfaces. The lower end is connected to the wall by a horizontal rope that is frayed and can support a maximum tension of only 80.0 N.

PHY 113 C Fall 2013 -- Lecture 17