1

Physics 140 – Winter 2014April 22

Final Sampling of Problems, Principles and Techniques

Reminders:• Final exam Friday at 7:30 p.m. – arrive at least 5 minutes early• Two-hour exam – 25 questions – Covers chapters 1-15 • Last names starting in A-J: 170 Dennison (next door)• Last names starting in K-Z: 182 Dennison (here)• Four note cards (or one 8.5”x11” paper) allowed – otherwise rules

are same as before• Please sit in alternating columns

2

Multiple oscillations Not critically damped Not overdamped

Oscillation amplitude decreasing Damped

3

4

U + K =Mgy+12

Mv2 +12

ICMω2 =Constant = Mgy0

R a, v ICM ≡λMR2

Method 1 – energy conservation:

→1

2Mv2 +

1

2λ MR2( )

v

R⎛⎝⎜

⎞⎠⎟

2

= Mgy0 −Mgy

→1

2(1 + λ )v2 = g(y0 − y)

→ v = 2g(y0 − y)

1+ λ

Assume all start at height y0

Fastest average speed for smallest λ

Solid sphere (λ=2/5) wins

τ∑ =MgRsin(θ ) = (ICM +MR2 )α =(λ MR2 +MR2 )a

R⎛⎝⎜

⎞⎠⎟

5

a, v

Method 2 – dynamics – define torque origin at contact point

Greatest acceleration for smallest λ

Solid sphere (λ=2/5) wins→ a =

gsin(θ )

1+ λ ✔

mg

Lever arm of weight: Rsin(θ)

6

Fully inelastic collision: mbulletvbullet =mbothvboth → vboth =mbullet

mboth

vbullet

Time t to fall computed from: H =12

gt2 → t =2H

g

Horizontal distance traveled during fall: D =v0xt=vbotht

→ D =.015

.815(400 m/s)

2(0.8 m)

9.8 m/s2= 2.97 m

M in =Vinρ =VinMM

VM

F =GmM in

r2 =Gmr2

r3MM

RM3 =

GmMM

RM3

r

=r3MM

RM3

→ "k" ≡GmMM

RM3

(SHM)

Potential energy of mass m at radius r:

U(r) =12

kr2

Potential energy of mass at surface relative to center:

U =12

kRM2 =

12

GmMM

RM3 RM

2 =12

GmMM

RM

Converted to kinetic energy at center:

K =12

mvmax 2 =

12

GmMM

RM

→ vmax =GMM

RM

Alternative shortcut – for any SHM:

vmax =ωA=km

A ✔

vmax =(6.67 ×10−11 Ngm2 / kg2 )(7.35 ×1022 kg)

1.74 ×106 m = 1.68 km/s

=GmMM

R3Mm

RM =GMM

RM

9

10

Mg

Mg

θ

τ∑ =−(Mg)L

2sin(θ )

⎛⎝⎜

⎞⎠⎟

− (Mg) L sin(θ )( )

= Ipivotα = Ipivot

d 2θ

dt 2

d 2θdt2

=−32

MgLIpivot

sin(θ)

=−3

2

MgL

ML

2⎛⎝⎜

⎞⎠⎟

2

+ML2

sin(θ )

=−(3)(4)MgL

(2)(5)ML2sin(θ ) ≅ −

6

5

g

Lθ

→ ω 2 =6

5

g

L→ T =

2π

ω= 2π

5

6

L

g

Lsin(θ)

11

ume

m1gm2g

a

TT

Forces on m1: T −m1g=m1a

Forces on m2 : m2g−T =m2a+ ++

→ m2g − m1g = m1a + m2a

=6 − 3

3 + 6g =

g

3

T =m1(g+ a) =43

m1g

=4

3(3 kg)(9.8 m/s2 )

= 39.2 N

→ a =m2 − m1

m1 + m2

g

12

T

TF

F

E

13τ∑ =M (4 cm) − (1.5 kg)(9.8 m/s2 )(14 cm) − (8.0 kg)(9.8 m/s2 )(30 cm) = 0