PHYSICS 231

INTRODUCTORY PHYSICS I

PHYSICS 231

INTRODUCTORY PHYSICS I

Lecture 12

• Newton’s Law of gravitation

• Kepler’s Laws of Planetary motion

1. Ellipses with sun at focus

2. Sweep out equal areas in equal times

3.

Last Lecture

€

Fgrav =GMm

R2

G = 6.67 ⋅10−11 Nm2 /kg2

€

R3

T 2=GM

4π 2= Constant

• PE = mgh valid only near Earth’s surface

• For arbitrary altitude

• Zero reference level isat r=

PE =−GMmr

Gravitational Potential Energy

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

Rotational Equilibrium and

Rotational Dynamics

Wrench Demo

Torque

• Torque, , 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

=Fd

Door Demo

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 FF

- +

Non-perpendicular forces

Φ is the angle between F and r

=Fr sinφ

Torque and Equilibrium

• Forces sum to zero (no linear motion)

• Torques sum to zero (no rotation)

ΣFx = 0 and ΣFy = 0

Σ =0

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

Given M = 120 kg.Neglect the mass of the beam.

a) Find the tensionin the cable

b) What is the forcebetween the beam andthe wall

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

Another Example

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

Find the tension in the muscle

F = 583 N

xL

W

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: the center-of-gravity.

• Center of gravity: €

= (mig)x i∑ = M totgmix i∑M tot

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟= M totgX

€

X =mix i∑M tot

Example 8.2

Fig 8.12, p.228

Slide 17

Given: x = 1.5 m, L = 5.0 m, wbeam = 300 N, wman = 600 NFind: T

x

L

T = 413 N

Example 8.3

Consider the 400-kgbeam shown below.Find TR

TR = 1 121 N

Example 8.4a

Given: Wbeam=300 Wbox=200Find: Tleft

Wbeam

Tleft Tright

Wbox

8 m2 m

A B C D

What point should I use for torque origin?ABCD

Example 8.4b

Given: Tleft=300 Tright=500Find: Wbeam

Wbeam

Tleft Tright

Wbox

8 m2 m

A B C D

What point should I use for torque origin?ABCD

Example 8.4c

Given: Tleft=250 Tright=400Find: Wbox

Wbeam

Tleft Tright

Wbox

8 m2 m

A B C D

What point should I use for torque origin?ABCD

Example 8.4d

Given: Wbeam=300 Wbox=200Find: Tright

Wbeam

Tleft Tright

Wbox

8 m2 m

A B C D

What point should I use for torque origin?ABCD

Example 8.4e

Given: Tleft=250 Wbeam=250Find: Wbox

Wbeam

Tleft Tright

Wbox

8 m2 m

A B C D

What point should I use for torque origin?ABCD

F =ma, =Iα

Torque and Angular Acceleration

Analogous to relation between F and a

Moment of Inertia

m

F

R

€

=FR = (mat )R = m(αR)R

τ = mR2α

Moment of Inertia

• Moment of inertia, I: rotational analog to mass

• r defined relative to rotation axis• SI units are kg m2

I = miri2

i∑

Baton DemoMoment-of-Inertia Demo

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

• Divide ring into segments

• The radius of each segment is R

I =Σmiri2 =MR2

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 kgm2

b) 1.08 kgm2

c) 1.8 kgm2

Other Moments of Inertia

Other Moments of Inertia

bicycle rim

filled can of coke

baton

baseball bat

basketball

boulder

cylindrical shell : I =MR2

solid cylinder : I =12MR2

rod about center : I =112

ML2

rod about end: I =13ML2

spherical shell : I =23MR2

solidsphere: I =25MR2

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 kgm2 b) 6.67 rad/s2 c) 4 m/s2 d) 1.72 kg

Example 8.9

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