GENERAL PHYSICS PH 221-1D (Dr. S. Mirov) Test 3 (04/05/13)

STUDENT NAME: ________________________ STUDENT id #: ___________________________ -------------------------------------------------------------------------------------------------------------------------------------------

ALL QUESTIONS ARE WORTH 25 POINTS. WORK OUT FOUR PROBLEMS. NOTE: Clearly write out solutions and answers (circle the answers) by section for each part (a., b., c., etc.)

Important Formulas:

1. Motion along a straight line with a constant acceleration vaver. speed = [dist. taken]/[time trav.]=S/t; vaver.vel. = ∆x/∆t; vins =dx/∆t; aaver.= ∆vaver. vel./∆t; a = dv/∆t; v = vo + at; x= 1/2(vo+v)t; x = vot + 1/2 at2; v2 = vo

2 + 2ax (if xo=0 at to=0)

2. Free fall motion (with positive direction ↑) g = 9.80 m/s2; y = vaver. t vaver.= (v+vo)/2; v = vo - gt; y = vo t - 1/2 g t2; v2 = vo

2 – 2gy (if yo=0 at to=0)

3. Motion in a plane vx = vo cosθ; vy = vo sinθ; x = vox t+ 1/2 ax t2; y = voy t + 1/2 ay t2; vx = vox + at; vy = voy + at;

4. Projectile motion (with positive direction ↑) vx = vox = vo cosθ; x = vox t; xmax = (2 vo

2 sinθ cosθ)/g = (vo2 sin2θ)/g for yin = yfin;

vy = voy - gt = vo sinθ - gt; y = voy t - 1/2 gt2;

5. Uniform circular Motion a=v2/r, T=2πr/v

6. Relative motion PA PB BA

PA PB

v v va a

= +=

7. Component method of vector addition

key

A = A1 + A2 ; Ax= Ax1 + Ax2 and Ay = Ay1 + Ay2; A A Ax y= +2 2 ; θ = tan-1 Ay /Ax;

The scalar product A = cosa b ab φ⋅

ˆ ˆˆ ˆ ˆ ˆ( ) ( )x y z x y za b a i a j a k b i b j b k⋅ = + + ⋅ + +

= x x y y z za b a b a b a b⋅ + +

The vector product ˆ ˆˆ ˆ ˆ ˆ( ) ( )x y z x y za b a i a j a k b i b j b k× = + + × + +

ˆˆ ˆˆˆ ˆ

ˆˆ ˆ( ) ( ) ( )

y z x yx zx y z

y z x yx zx y z

y z y z z x z x x y x y

i j ka a a aa a

a b b a a a a i j kb b b bb b

b b b

a b b a i a b b a j a b b a k

× = − × = = − + =

= − + − + −

1. Second Newton’s Law ma=Fnet ;

2. Kinetic friction fk =µkN;

3. Static friction fs =µsN;

4. Universal Law of Gravitation: F=GMm/r2; G=6.67x10-11 Nm2/kg2;

5. Drag coefficient 212

D C Avρ=

6. Terminal speed 2

tmgv

C Aρ=

7. Centripetal force: Fc=mv2/r

8. Speed of the satellite in a circular orbit: v2=GME/r

9. The work done by a constant force acting on an object: cosW Fd F dφ= = ⋅

10. Kinetic energy: 212

K mv=

11. Total mechanical energy: E=K+U

12. The work-energy theorem: W=Kf-Ko; Wnc=∆K+∆U=Ef-Eo

13. The principle of conservation of mechanical energy: when Wnc=0, Ef=Eo

14. Work done by the gravitational force: cosgW mgd φ=

1. Work done in Lifting and Lowering the object:

; ; f i a g f i a gK K K W W if K K W W∆ = − = + = = −

2. Spring Force: (Hook's law)xF kx= −

3. Work done by a spring force: 2 2 21 1 1; if 0 and ; 2 2 2s i o i f sW kx kx x x x W kx= − = = = −

4. Work done by a variable force: ( )f

i

x

x

W F x dx= ∫

5. Power: ; ; cosavgW dWP P P Fv F v

t dtφ= = = = ⋅

∆

6. Potential energy: ; ( )f

i

x

xU W U F x dx∆ = − ∆ = −∫

7. Gravitational Potential Energy:

( ) ; 0 and 0; ( )f i i iU mg y y mg y if y U U y mgy∆ = − = ∆ = = =

8. Elastic potential Energy: 21( )2

U x kx=

9. Potential energy curves: ( )( ) ; ( ) ( )mec

dU xF x K x E U xdx

= − = −

10. Work done on a system by an external force:

Friction is not involved When kinetic friction force acts within the system

mec

mec th

th k

W E K UW E E

E f d

= ∆ = ∆ + ∆= ∆ + ∆

∆ =

11. Conservation of energy: int

intfor isolated system (W=0) 0mec th

mec th

W E E E EE E E

= ∆ = ∆ + ∆ + ∆∆ + ∆ + ∆ =

12. Power: ; avgE dEP Pt dt

∆= =∆

;

13. Center of mass: 1

1 n

com i ii

r m rM =

= ∑

14. Newtons’ Second Law for a system of particles: net comF Ma=

1. Linear Momentum and Newton’s Second law for a system of particles: and com netdPP Mv Fdt

= =

2. Collision and impulse: ( ) ; ;

f

i

t

avgtJ F t dt J F t= = ∆∫

when a stream of bodies with mass m and

speed v, collides with a body whose position is fixed avgn n mF p m v vt t t

∆= − ∆ = − ∆ = − ∆

∆ ∆ ∆

Impulse-Linear Momentum Theorem: f ip p J− =

3. Law of Conservation of Linear momentum: for closed, isolated systemi fP P=

4. Inelastic collision in one dimension: 1 2 1 2i i f fp p p p+ = +

5. Motion of the Center of Mass: The center of mass of a closed, isolated system of two colliding bodies is

not affected by a collision.

6. Elastic Collision in One Dimension: 1 2 11 1 2 1

1 2 1 2

2; f i f im m mv v v vm m m m

−= =

+ +

7. Collision in Two Dimensions: 1 2 1 2 1 2 1 2; ix ix fx fx iy iy fy fyp p p p p p p p+ = + + = +

8. Variable-mass system:

(first rocket equation)

ln (second rocket equation)

rel

if i rel

f

Rv MaMv v vM

=

− =

9. Angular Position: (radian measure)Sr

θ =

10. Angular Displacement: 2 1 (positive for counterclockwise rotation)θ θ θ∆ = −

11. Angular velocity and speed: ; (positive for counterclockwise rotation)avgd

t dtθ θω ω∆

= =∆

12. Angular acceleration: ; avgd

t dtω ωα α∆

= =∆

1. angular acceleration: 2

2 2

2

1 ( )2

12

2 ( )12

o

o o

o o

o o

o

t

t

t t

t t

ω ω α

θ θ ω ω

θ θ ω ω

ω ω α θ θ

θ θ ω ω

= +

− = +

− = +

= + −

− = −

2. Linear and angular variables related:

2

2 2 2; ; ; ; t rv rs r v r a r a r Tr v

π πθ ω α ωω

= = = = = = =

3. Rotational Kinetic Energy and Rotational Inertia:

2 2

2

1 ; for body as a system of discrete particles;2

for a body with continuously distributed mass.

i iK I I m r

I r dm

ω= =

=

∑

∫

4. The parallel axes theorem: 2comI I Mh= +

5. Torque: sintrF r F rFτ φ⊥= = =

6. Newton’s second law in angular form: net Iτ α=

7. Work and Rotational Kinetic Energy: ; ( ) for ;f

if id W constW

θ

θτ θ τ θ θ τ= − == ∫

2 21 1; work energy theorem for rotating bodies2 2f i f i

dWP K K K I I Wdt

ω ω= ∆ = − = − =

8. Rolling bodies:

2 2

2

1 12 2

sin for rolling smoothly down the ramp1 /

com

com com

com

comcom

v R

K I mv

a RgaI MR

ω

ω

αθ

=

= +

=

=+

9. Torque as a vector: ; sinr F rF rF r Fτ τ φ ⊥ ⊥= × = = =

1. Angular Momentum of a particle: ( );

sinl r p m r vl rmv rp rmv r p r mvφ ⊥ ⊥ ⊥ ⊥

= × = ×= = = = =

2. Newton’s Second law in Angular Form: netdldt

τ =

3. Angular momentum of a system of particles: 1

n

ii

net ext

L l

dLdt

τ

=

=

=

∑

4. Angular Momentum of a Rigid Body: L Iω=

5. Conservation of Angular Momentum: (isolated system)i fL L=

6. Static equilibrium: net

, , ,

0; 0if all the forces lie in xy plane 0; 0; 0

net

net x net y net z

FF F

ττ

= == = =

7. Elastic Moduli: stress=modulus strain×

8. Tension and Compression: , E is the Young's modulusF LEA L

∆=

9. Shearing: , G is the shear modulusF LGA L

∆=

10. Hydraulic Stress: , B is the bulk modulusVp BV∆

=

7

1. Three identical rods are rigidly connected to a disk, as shown in Fig. Find the center of mass of the system.

20 cm 20 cm

R=8.0 cm m1=40 g

m2=40 g

m4=20 g

20 cm

m3=40 g

x

y

Considering the center of the disk as (0,0) :( 18)(40) (18)(40) 0 0 0;

1400 0 0 (18)(40) 5.1

140

com

com

x

y cm

− + + += =

+ + += =

2. A 640-N hunter gets a rope around a 3200-N polar bear. They are stationary, 20m apart, on frictionless level ice. When the hunter pulls the polar bear to him, how far does the polar bear travel?

0 +x

20m 640N 3200N

1 1 2 2 1 1 2 2

1 2 1

( ) Since hunter-polar bear represent an isolated system and initially they are at resttheir center of mass will be at rest and they will meet at the center of mass

com

a

m x m x m gx m gxxm m m g m

+ += =

+ +(640)(0) (3200)(20)

640 32002

16.7

(b) The 640 N hunter moves 16.7 m(c) The 3200 N polar bear moves 20-16.7=3.3m

mg

++= =

3. Blocks A and B are moving toward each other. A has a mass of 2.0kg and a velocity of 50m/s, while B has a mass of 4.0kg and a velocity of -25m/s. They suffer a completely inelastic collision. What is the kinetic energy lost during the collision?

mA mB

v10=50 m/s V20=-25m/s

x

1 10 2 20

(a) Consider inelastic block-block collision. At the instant of collision block-block system isisolated, hence, we can use law of conservation of linear momentum to describe the collision

(m v m v− = 1 10 2 201 2

1 2

2 2 2 21 1 2 2

(2.0)(50) (4.0)(25)) ; 0.0 /( ) (2.0 4.0)

(b) Hence, 0.

(c) Calculate final kinetic energy. 2.0 50 4.0 ( 25) 2500 1250 3750

2 2 2 2(d) Kinetic energy lost dur

f

f

m v m vm m v v m sm m

K

m v m vK J

− −+ ⇒ = = =

+ +

=

⋅ ⋅ −= + = + = + =

3

ing the collision 3750 0 3750 3.8 10i fK K K J J∆ = − = − = = ×

10

4.

11

5. In Figure below a solid ball ( I MR= 25

2 ) rolls smoothly from rest (starting at height H=6.0 m) until it leaves the horizontal section at the end of the track, at height h=2.0 m. How far horizontally from point A does the ball hit the floor?

To find where the ball lands, we need to know its speed as it leaves the track (using conservation of energy). Its initial kinetic energy is Ki = 0 and its initial potential energy is Ui = M gH. Its final kinetic energy (as it leaves the track) is K Mv If = +1

22 1

22ω and its

final potential energy is M gh. Here we use v to denote the speed of its center of mass and ω is its angular speed — at the moment it leaves the track. Since (up to that moment) the ball rolls without sliding we can set ω = v/R. Using I MR= 2

52 , conservation of energy

leads to

2 2 2 2 21 1 1 2 7 .2 2 2 10 10

MgH Mv I Mgh Mv Mv Mgh Mv Mghω= + + = + + = +

The mass M cancels from the equation, and we obtain

v g H h= − = − =107

107

9 8 6 0 2 0 7 482b g d ib g. . . . .m s m m m s

Now this becomes a projectile motion. We put the origin at the position of the center of mass when the ball leaves the track (the “initial” position for this part of the problem) and take +x rightward and +y downward. Then (since the initial velocity is purely horizontal) the projectile motion equations become

x vt y gt= = −and 12

2.

Solving for x at the time when y = h, the second equation gives t h g= 2 . Then, substituting this into the first equation, we find

( ) ( )2

2 2.0 m2 7.48 m/s 4.8 m.9.8 m/s

hx vg

= = =

6. A playground merry-go-round has a radius of 3.0m and a rotational inertia of 600 kg⋅m2. It is initially spinning at 0.80 rad/s when a 20-kg child crawls from the center to the rim. When the child reaches the rim what is the angular velocity of the merry-go-round?

a) Playground-child represent an isolated system, because the net external torqueabout the playground axis of rotaion is zero. Hence, one can use law of conservation ofangular momentum to solve the pr

2 2

2 2

2 2

oblem.

b) ; ( ) ( )

( ) (600 20 0 ) 0.8 0.62 /( ) 600 20 3.0

i f p c i o p c f f

p c i of

p c f

L L I m r I m r

I m rrad s

I m r

ω ω

ωω

= + = +

+ + ⋅ ⋅= = =

+ + ⋅

13

7. A uniform horizontal beam of length 8.00 m and weight 200 N is attached to a wall by a pin connection. Its far end is supported by a cable that makes an angle of 53.0° with the horizontal. If a 600 N man stands 2.00 m from the wall, find the tension in the cable and the force exerted by the wall on the beam.