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PHYSICS 231 INTRODUCTORY PHYSICS I. Lecture 12. Last Lecture. Newton’s Law of gravitation Kepler’s Laws of Planetary motion Ellipses with sun at focus Sweep out equal areas in equal times. Gravitational Potential Energy. PE = mgh valid only near Earth’s surface For arbitrary altitude

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Physics 231 introductory physics i

PHYSICS 231INTRODUCTORY PHYSICS I

Lecture 12


Last lecture
Last Lecture

  • Newton’s Law of gravitation

  • Kepler’s Laws of Planetary motion

    • Ellipses with sun at focus

    • Sweep out equal areas in equal times


Gravitational potential energy
Gravitational Potential Energy

  • PE = mgh valid only near Earth’s surface

  • For arbitrary altitude

  • Zero reference level isat r=


Example 7 18
Example 7.18

You wish to hurl a projectile from the surface of the Earth (Re= 6.38x106 m) to an altitude of 20x106 m above the surface of the Earth. Ignore rotation of the Earth and air resistance.

a) What initial velocity is required?

b) What velocity would be required in order for the projectile to reach infinitely high? I.e., what is the escape velocity?

c) (skip) How does the escape velocity compare to the velocity required for a low earth orbit?

a) 9,736 m/s

b) 11,181 m/s

c) 7,906 m/s


Chapter 8

Chapter 8

Rotational Equilibrium

and

Rotational Dynamics



Torque
Torque

  • Torque, t , is tendency of a force to rotate object about some axis

    • F is the force

    • d is the lever arm (or moment arm)

    • Units are Newton-meters

Door Demo


Torque is vector quantity
Torque is vector quantity

  • Direction determined by axis of twist

  • Perpendicular to both r and F

  • Clockwise torques point into paper. Defined as negative

  • Counter-clockwise torques point out of paper. Defined as positive

r

r

F

F

+

-


Non perpendicular forces
Non-perpendicular forces

Φ is the angle between F and r


Torque and equilibrium
Torque and Equilibrium

  • Forces sum to zero (no linear motion)

  • Torques sum to zero (no rotation)


Axis of rotation
Axis of Rotation

  • Torques require point of reference

  • Point can be anywhere

    • Use same point for all torques

    • Pick the point to make problem easiest (eliminate unwanted Forces from equation)


Example 8 1
Example 8.1

Given M = 120 kg.

Neglect the mass of the beam.

a) Find the tensionin the cable

b) What is the force

between the beam andthe wall

a) T=824 N b) f=353 N


Another example
Another Example

Given: W=50 N, L=0.35 m, x=0.03 m

Find the tension in the muscle

W

x

L

F = 583 N


Center of gravity
Center of Gravity

  • Gravitational force acts on all points of an extended object

  • However, one can treat gravity as if it acts at one point: thecenter-of-gravity.

  • Center of gravity:


Example 8 2
Example 8.2

Given: x = 1.5 m, L = 5.0 m, wbeam = 300 N,

wman = 600 N

Find: T

T = 413 N

x

L


Example 8 3
Example 8.3

Consider the 400-kgbeam shown below.Find TR

TR = 1 121 N


Example 8 4a
Example 8.4a

Given:

Wbeam=300

Wbox=200

Find:

Tleft

A

B

C

D

What point should I use for torque origin?


Example 8 4b
Example 8.4b

Given:

Tleft=300

Tright=500

Find:

Wbeam

A

B

C

D

What point should I use for torque origin?


Example 8 4c
Example 8.4c

Given:

Tleft=250

Tright=400

Find:

Wbox

A

B

C

D

What point should I use for torque origin?


Example 8 4d
Example 8.4d

Given:

Wbeam=300

Wbox=200

Find:

Tright

A

B

C

D

What point should I use for torque origin?


Example 8 4e
Example 8.4e

Given:

Tleft=250

Wbeam=250

Find:

Wbox

A

B

C

D

What point should I use for torque origin?


Torque and angular acceleration

F

m

R

Torque and Angular Acceleration

Analogous to relation between F and a

Moment of Inertia


Moment of inertia
Moment of Inertia

  • Moment of inertia, I: rotational analog to mass

    • r defined relative to rotation axis

    • SI units are kg m2


Baton demo moment of inertia demo
Baton DemoMoment-of-Inertia Demo


More about moment of inertia
More About Moment of Inertia

  • Depends on mass and its distribution.

  • If mass is distributed further from axis of rotation, moment of inertia will be larger.


Moment of inertia of a uniform ring
Moment of Inertia of a Uniform Ring

  • Divide ring into segments

  • The radius of each segment is R


Example 8 6
Example 8.6

What is the moment of inertia of the following point masses arranged in a square?

a) about the x-axis?

b) about the y-axis?

c) about the z-axis?

a) 0.72 kgm2

b) 1.08 kgm2

c) 1.8 kgm2



Other moments of inertia1
Other Moments of Inertia

bicycle rim

filled can of coke

baton

baseball bat

basketball

boulder


Example 8 7
Example 8.7

Treat the spindle as a solid cylinder.

a) What is the moment of Inertia of the spindle? (M=5.0 kg, R=0.6 m)

b) If the tension in the rope is 10 N, what is the angular acceleration of the wheel?

c) What is the acceleration of the bucket?

d) What is the mass of the bucket?

M

a) 0.9 kgm2 b) 6.67 rad/s2 c) 4 m/s2 d) 1.72 kg


Example 8 9
Example 8.9

A 600-kg solid cylinder of radius 0.6 m which can rotate freely about its axis is accelerated

by hanging a 240 kg mass from the end by a string which is wrapped about the cylinder.a) Find the linear acceleration of the mass.

b) What is the speed of the mass after it has dropped 2.5 m?

4.36 m/s2

4.67 m/s


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