lecture 24 – pendula and gravity today’s...

Lecture 24 – Pendula and Gravity

Week 13 Assignments: (Due 11/17 {Tuesday} by the end of class) Textbook: HW #10 Chp 14: Q1, Q7, Q11, P4, P6, P13[B only], P17, P41, P43, P55

MasteringPhysics: - Assignment 10 (Due 11/17 {Tuesday} by the end of class)

Week 13 Reading: Chapter 6.1-6.6 & 8.7 -- Giancoli

Today’s Announcements: * Midterm available to pick up in recitation

* Midterm extra credit (due the last regular thursday of class):

- MidtermEC_1 (if midterm I is lowest score)

- MidtermEC_2 (if midterm II is lowest score)

- you must do the problems by yourself (no working in groups, no consulting the TA, internet)

- though you may ask me questions / clarifications

- 1 problem = 1 midterm problem (a maximum improvement of your midterm score of 10% is possible)

- Open book / Open notes / open calculator

Pendulum

mg sinθ

- τ = I α = r x F

⇒ I α = - r (mg) sinθ

I d2θ/dt2 = -Lmg sinθ

ALMOST the second derivative of “something” equals a constant (-Lmg/ I) times that “something”, but not quite: [sin(“something”)]

sinθ ≈ θ when θ is small (<20o or so)

d2θ/dt2 ≈ -(Lmg/I) θ - when oscillations are small

Same math form as SHO --- same form of solution

θ L

mg

T

Simple Pendulum

d2θ/dt2 ≈ -(Lmg/I) θ

- Try the basic solution: θ(t) = A cos(ω t + φ)

⇒ - ω2 A cos(ω t + φ) = (-Lmg/I) A cos(ω t + φ)

- Same exact thing as the spring except that: x(t) ⇒ θ(t) and k ⇒ mgL = κ and m ⇒ I !

ω2 = mgL/I

- Simple Pendulum: I = mL2!ω2 = g/L

T = 2π(L/g)0.5

mg sinθ

θ L

mg

T

Clicker Question:

a) Pluto, with g =0.4 m/s2

6)

b) The Moon, with g =1.62 m/s2

c) Neptune’s moon Triton, with g =2.5 m/s2

d) Mars, with g =3.74 m/s2

e) Venus, with g =8.87 m/s2

T = 1.62 s

L = 0.25m

T = 2π(L/g)0.5

You wake up, finding yourself abducted by space aliens. You have no idea where you are, but you are a smart Physics 121 grad. The aliens have hung you upside down by your feet (ala The Empire Strikes Back). But you happen to be wearing a necklace (of length 0.25 m) with a heavy medallion on the end. You swing the necklace and see that it takes T = 1.62 s to complete a full cycle. Where are you ?

Clicker Question:

a) Pluto, with g =0.4 m/s2

6)

b) The Moon, with g =1.62 m/s2

c) Neptune’s moon Triton, with g =2.5 m/s2

d) Mars, with g =3.74 m/s2

e) Venus, with g =8.87 m/s2

T = 1.62 s

L = 0.25m

T = 2π(L/g)0.5

You wake up, finding yourself abducted by space aliens. You have no idea where you are, but you are a smart Physics 121 grad. The aliens have hung you upside down by your feet (ala The Empire Strikes Back). But you happen to be wearing a necklace (of length 0.25 m) with a heavy medallion on the end. You swing the necklace and see that it takes T = 1.62 s to complete a full cycle. Where are you ?

Physical Pendulum

ω2 = mgL/I

- Physical Pendulum:!

L gets replaced by Rcm

T = 2π(CL2/gRcm)0.5

(Iobj = CmL2) !

-  Angular frequency is still the same: !

- But: I gets replaced by Iobj c.m. ×

Rcm L

- τ = I α = R x F

⇒ I α = -Rcmmgsinθ

θ

mg

Chp 13:

- So far our discussion of gravity has been confined to near the surface of the earth where we have represented gravity just by an acceleration, g. This week we wish to give a more general representation of gravity.

Galileo and Falling Bodies

- Galileo demonstrated that falling bodies hit the ground at the same time independent of their mass

tff = (2h/ge)0.5

Galileo and Falling Bodies

- Galileo demonstrated that falling bodies hit the ground at the same time independent of their mass

tff = (2h/ge)0.5

Johannes Kepler (ca. 1609):

Meanwhile Back in Prague…

Discovers 3 Laws of planetary motion:

v2

m R2

v1 m

R1

A1

A2

+

+

foci 2a

T2 = C a3 with C = same constant for all planets

- Newton attempts to incorporate Galileo’s falling body data, Kepler’s planetary motions and his own three laws of motion into a coherent description of gravity

Isaac Newton

Gravitational Force

Fgrav1

Fgrav ∝ m = ma m

- Because all objects accelerate at the same rate, Fgrav must depend linearly on the mass of the falling object (Newton’s 2nd Law)

Fgrav2

Fgrav ∝ Mem

Me

- If however we look at it from the perspective of an ant on the ball, you must rightfully say that the earth is falling down to the ball (Newton’s 3rd Law), so the Fgrav must depend linearly on the other mass |Fgrav1| = |Fgrav2|

Fgrav ∝1/R2center

- From calculus of ellipses:

Which Newton (co-)invented

•  Any two masses have a gravitational force between them:

•  M1 and M2 are the masses •  R is the distance between the centers

of the two masses (direction = attractive) •  G is the gravitational constant

(G = 6.67 x 10-11 N kg2/m2)

M1

M2

R

Acceleration due to Gravity, g

F = GMemo/R2 Me

mo

R

F = mog

= GMemo/R2

- So g is related to the size and mass of the earth !

g = GMe/R2

g

- If you know G and Re then g tells you the mass of the earth (“weighing the earth”)

- So how does this relate to g?

Before (chp 4-8):