Physics 2210 Fall 2015

smartPhysics 02-03 02 Projectile Motion (continued)

03 Relative Motion, Circular Motion 09/02/2015

Events Dates

Labor Day holiday Monday, September 7

Tuition payment due Friday, September 4

Classes begin Monday, August 24

Last day to add without a permission code Sunday, August 30

Last day to add, drop (delete), elect CR/NC, or audit classes

Friday, September 4

Last day to withdraw from classes Friday, October 23

Tuesdays 8:30 - 9:20 am…………..JWB 308 Wednesdays 3:00 - 3:50 pm………JFB B-1 Thursdays 11:30am - 12:20 pm…...LCB 121

Supplemental Instruction Schedule

Reminder of Important Dates

(politically incorrect) Example 2.2 (1/4) From ground level, a hunter shoots at a monkey hanging in a tree. The monkey is originally at height h and distance D from the hunter. When the gun is fired, the monkey hears the shot instantaneously (assume sound travels very fast), lets go of the tree branch, and falls. (a) Treating h, D, and v0 (initial velocity of the bullet) as given parameters, at what angle θ above the ground should the hunter aim in order to hit the monkey? (b) What is the minimum v0 required for success?

y

x O

h

D

v0

θ

Monkey starts dropping vertically at t = 0

Gun fired from origin at t = 0

(politically incorrect) Example 2.2 (2/4) The monkey is originally at height h and distance D from the hunter. When the gun is fired, the monkey starts dropping. At what angle θ above the ground should the hunter aim in order to hit the monkey (%i1) /* The bullet travels according to projectile motion: constant

velocity in the x-direction (horizontal, right+) and constant acceleration at aBy=-g in the y (vertical, up+) direction. Staring at t=0 from the origin: xB0=0 and yB0=0, the bullet's position is given by */

xB: vBx0*t;

(%o1) t vBx0

(%i2) yB: vBy0*t - g/2*t^2;

2

g t

(%o2) t vBy0 - ----

2

(%i3) /* The monkey starts at initial position (D, h) from rest, so its position is given by */

xM: D;

(%o3) D

(%i4) yM: h - g/2*t^2;

2

g t

(%o4) h - ----

2

(politically incorrect) Example 2.2 (3/4) The monkey is originally at height h and distance D from the hunter. When the gun is fired, the monkey starts dropping. (a) At what angle θ above the ground should the hunter aim in order to hit the monkey? (%i5) /* First find time t1 at which the bullets reaches the horizontal position of the monkey */

t1: rhs(solve(xB = xM, t)[1]);

D

(%o5) ----

vBx0

(%i6) /* at t=t1, the bullet and monkey are at height yB1 and yM1, respectively, given by */

yB1: yB, t=t1;

2

D vBy0 g D

(%o6) ------ - -------

vBx0 2

2 vBx0

(%i9) sol2: solve(yB1 = yM1,vBy0); h vBx0

(%o9) [vBy0 = ------]

D

Answer (a) vBy0/vBx0 = h/D = tan(theta)… it means you aim right at the monkey i.e. theta = arctan(h/D)

(politically incorrect) Example 2.2 (3/4) The monkey is originally at height h and distance D from the hunter. When the gun is fired, the monkey starts dropping. (a) At what angle θ above the ground should the hunter aim in order to hit the monkey? (b) What is the minimum v0 required for success? (%i10) /* we want to make sure yB1 > 0 *, so we solve for vBy0 such

that yB1=0 as the limiting case */

sol3: solve(yB1=0, vBy0);

g D

(%o10) [vBy0 = ------]

2 vBx0

So the requirement is that

2*vBy0*vBx0 > g*D

But vBx0 = v0*cos(theta)

And vBy0 = v0*sin(theta)

where theta is NOW the optimized angle given by tan(theta)=h/D

And 2*sin(theta)*cos(theta) = sin(2*theta)

So the minimum v0 is given by

v0_min = sqrt( g*D/sin(2*theta) ) = sqrt( g*D/(2*sin(theta)*cos(theta)) )

24 fps https://www.youtube.com/watch?v=0jGZnMf3rPo

Shoot the Monkey Demo

Special Conditions during Projectile Motion Instead of memorizing the formula for H and R, It is more powerful and useful to UNDERSTANF what it means to reach maximum height and hitting ground again

𝑣𝑥 = 𝑣𝑥𝑥 𝑥 = 𝑥𝑥 + 𝑣𝑥𝑥𝑡 𝑣𝑦 = 𝑣𝑦𝑥 − 𝑔𝑡 𝑦 = 𝑦𝑥 + 𝑣𝑦𝑥𝑡 − 1

2𝑔𝑡2

t = 0 (𝒙𝟎,𝒚𝟎) 𝒚 = 𝒚𝟎

Unit 03

Unit 03

Unit 3: (A) Relative Velocities Object B moves at velocity �⃗�𝐵𝐵 (usually abbreviate to �⃗�𝐵) relative to the “Lab frame”, and Object A at velocity �⃗�𝐴𝐵 (abbreviate to �⃗�𝐴) Then the velocity of B relative to A, �⃗�𝐵𝐴(as measured by A) is given by

�⃗�𝐵𝐴 = �⃗�𝐵𝐵 − �⃗�𝐴𝐵 𝑜𝑜 �⃗�𝐵𝐴 = �⃗�𝐵 − �⃗�𝐴 In Cartesian components

𝑣𝐵𝐴𝑥 = 𝑣𝐵𝑥 − 𝑣𝐴𝑥 𝑣𝐵𝐴𝑦 = 𝑣𝐵𝑦 − 𝑣𝐴𝑦 (𝑣𝐵𝐴𝑧 = 𝑣𝐵𝑧 − 𝑣𝐴𝑧) (cannot add polar, spherical, or cylindrical coordinates like this) Rearranging :

�⃗�𝐵 = �⃗�𝐵𝐴 + �⃗�𝐴 �⃗�𝐴 = �⃗�𝐵 − �⃗�𝐵𝐴 = �⃗�𝐴𝐵 + �⃗�𝐵

Note the velocity of A relative to B, �⃗�𝐴𝐵 is equal and opposite to �⃗�𝐵𝐴 �⃗�𝐴𝐵 = −�⃗�𝐵𝐴

Poll 09-02-01

A girl woman stands on a moving sidewalk (conveyor belt) that is moving to the right at a speed of 2 m/s relative to the ground. A dog runs on the belt toward the girl woman at a speed of 8 m/s relative to the belt. What is the speed of the dog relative to the ground?

A. 10 m/s B. 8 m/s C. 6 m/s

Example 3.1 (1/1) An airliner cruises at 560 mi/hr. It is to travel from airport A to airport B. Airport B is 3500 miles east of A (similar distance to New York – London) (a) A steady, 230 mi/hr wind is blowing to the east. Neglecting

acceleration during take-off and landing, and pretending the Earth is flat, how long does the trip take (usually listed for 7.5 hours)?

(b) repeat (a) for the return trip. (a) Outbound

�⃗�𝑝𝑝 = �⃗�𝑝𝑎 + �⃗�𝑎𝑝 = 560𝑚𝑚ℎ �̂� + 230

𝑚𝑚ℎ �̂� = 790

𝑚𝑚ℎ �̂�

∆𝑡 =∆𝑥𝑣𝑝𝑝𝑥

=3500𝑚𝑚790𝑚𝑚/ℎ = 4.43 ℎ

(b) Return trip: we are now flying west, but wind still blows east.

�⃗�𝑝𝑝 = �⃗�𝑝𝑎 + �⃗�𝑎𝑝 = −560𝑚𝑚ℎ �̂� + 230

𝑚𝑚ℎ �̂� = −330

𝑚𝑚ℎ �̂�

∆𝑡 =∆𝑥𝑣𝑝𝑝𝑥

=−3500𝑚𝑚−330𝑚𝑚/ℎ = 10.61 ℎ

http://www.youtube.com/watch?v=TCUHQ_-l6Qg

B52 bombers have (for a long time classified) feature of articulated landing gears that allow the wheels to point forward while the plane is rotated to offset the wind velocity

http://www.youtube.com/watch?v=la-hSjKP2TU

Relative velocities are especially important in navigation In strong “cross-winds”: airplanes have to point in a different direction than runway in order to land

We had �⃗�𝐵 = �⃗�𝐵𝐴 + �⃗�𝐴 Here: �⃗�𝑟 = �⃗�𝑝 + �⃗�𝑤 B: plane �⃗�𝑟: ground velocity �⃗�𝑝: air velocity (relative to air)

A: wind �⃗�𝑊: wind velocity (relative to ground)

Example 3.2 (1/2) You are traveling on an airplane. The velocity of the plane with respect to the air is 130 m/s due east. The velocity of the air with respect to the ground is 47 m/s at an angle of 30° west of due north. (a) What is the speed of the plane with respect to the ground?

(%i1) /* let east be +x and north be +y plane's velocity components relative to air are */ v_pa_x: +130; v_pa_y: 0; (%o1) 130 (%o2) 0 (%i3) /* 30 deg west of north is 30+90 deg = 120 deg CCW from +x */ phi: 120*%pi/180, numer; (%o3) 2.094395102393195 (%i4) /* speed of air is 47 m/s ... calculate components */ v_ag: 47; (%o4) 47 (%i5) v_ag_x: v_ag*cos(phi); (%o5) - 23.49999999999999 (%i6) v_ag_y: v_ag*sin(phi); (%o6) 40.70319397786862 (%i7) /* now calculate components of velocity of plane relatve to ground */ v_pg_x: v_pa_x + v_ag_x; (%o7) 106.5 (%i9) v_pg_y: v_pa_y + v_ag_y; (%o9) 40.70319397786862 (%i10) /* speed relatve to ground */ v_pg: sqrt(v_pg_x^2+v_pg_y^2); (%o10) 114.0131571354815 Answer: (a) 114 m/s

Example 3.2 (2/2) You are traveling on an airplane. The velocity of the plane with respect to the air is 130 m/s due east. The velocity of the air with respect to the ground is 47 m/s at an angle of 30° west of due north. (b) What is the heading of the plane with respect to the ground? (Let 0° represent due north, 90° represents due east) (c) How far east will the plane travel in 1 hour?

(%i12) /* phi angle of plane relative to ground */

phi_pg: atan2(v_pg_y, v_pg_x)*180/%pi, numer;

(%o12) 20.91633696405919

(%i13) /* but this is the angle north of east, we want the angle east of north, which is 90 deg - pho_pg */

90-phi_pg;

(%o13) 69.08366303594081

Answer (b) 69 deg east of north

(%i16) /* distance east in one hour = 3600 s */

Dt: 3600; Dx: v_pg_x*Dt;

(%o16) 3600

(%o17) 383400.0

Answer (c) 383.4 km

Poll 09-02-02

A girl twirls a rock on the end of a string in a horizontal circle above her head. The diagram illustrates how this looks from above. If the string breaks at the instant shown, which arrow best represents the path the rock will follow?

Uniform Circular Motion

𝑎𝑐 ≡ �⃗�𝑐 =𝑣2

𝑜

Centripetal Acceleration

x

y

O

s φ

r

𝑜

�⃗�

∆𝑣 Uniform Circular Motion: The speed is constant

𝑑𝑣𝑑𝑡

≡𝑑 �⃗�𝑑𝑡

= 0

But the direction of the velocity changes continuously:

�⃗� ≡𝑑�⃗�𝑑𝑡

= −𝑣2

𝑜�̂�

Uniform Circular Motion

x

y

O

s φ

r

𝑜

𝑣 = 𝑠/𝑡, 𝑠 = 𝑜𝜙 (𝜙 expressed in radians) 𝑣 = 𝑜𝜙/𝑡 𝜙 = 𝑣𝑡/𝑜

Position vector: 𝑜 = 𝑥𝚤̂ + 𝑦𝚥̂ = 𝑜 cos𝜙 𝚤̂ + 𝑜 sin𝜙 𝚥̂

= 𝑜 cos 𝑣𝑟𝑡 𝚤̂ + sin 𝑣

𝑟𝑡 𝚥̂ Velocity vector:

�⃗� ≡𝑑𝑜𝑑𝑡

=𝑑𝑑𝑡

𝑜 cos 𝑣𝑟𝑡 𝚤̂ + 𝑜 sin 𝑣

𝑟𝑡 𝚥̂

= 𝒗 − sin 𝑣𝑟𝑡 𝚤̂ + cos 𝑣

𝑟𝑡 𝚥̂

=𝑣𝑜−𝑦𝚤̂ + 𝑥𝚥̂

Acceleration vector:

�⃗� ≡𝑑�⃗�𝑑𝑡

= 𝑣𝑑𝑑𝑡

− sin 𝑣𝑟𝑡 𝚤̂ + cos 𝑣

𝑟𝑡 𝚥̂ =𝒗𝟐

𝒓−𝑐𝑜𝑠 𝑣

𝑟𝑡 𝚤̂ − 𝑠𝑚𝑠 𝑣𝑟𝑡 𝚥̂

= −𝑣2

𝑜𝑜𝑜

= −𝑣2

𝑜�̂�

�⃗�

�⃗�𝑐

𝑎𝑐 ≡ �⃗�𝑐 =𝑣2

𝑜

Centripetal Acceleration

If the speed v is allowed to vary, then the total acceleration also has a tangential component:

�⃗� =𝑑�⃗�𝑑𝑡

= �⃗�𝑟 + �⃗�𝑡

x

y

O

s φ

r

𝑜

�⃗�

�⃗�𝑐 �⃗�𝑡

𝑎𝑟 = −𝑣2

𝑜

Radial Component

𝑎𝑡 =𝑑𝑣𝑑𝑡

Tangential component

Components are NOT magnitudes…. They carry a sign to indicate left/right, CCW/CW, east/west, forward/backward

𝑎𝑐 ≡ �⃗�𝑐 =𝑣2

𝑜

Centripetal Acceleration

Non-Uniform Circular Motion

Extra Example 3.2A (2/4) (%i3) /* The intended heading is actually at phi_g=30 degrees

(35 deg north of east), so that vgy/vgx=tan(phi_g) */

eqn1: vgy/vgx=tan(phi_g);

sin(phi) vc

(%o3) ---------------- = tan(phi_g)

vw + cos(phi) vc

(%i4) soln1: solve(eqn1, phi);

tan(phi_g) vw + cos(phi) tan(phi_g) vc

(%o4) [sin(phi) = --------------------------------------]

vc

%i6) /* put in numbers */

deg35: 35*%pi/180, numer;

(%o6) 0.6108652381980153

(%i5) eqn2: soln1[1];

tan(phi_g) vw + cos(phi) tan(phi_g) vc

(%o5) sin(phi) = --------------------------------------

vc

(%i11) eqn3: expand(eqn2), phi_g=deg35, vw=230, vc=560, numer;

(%o11) sin(phi) = 0.7002075382097096 cos(phi) + 0.2875852389075594

Extra Example 3.2A (3/4) (%o11) sin(phi) = 0.7002075382097096 cos(phi) + 0.2875852389075594

(%i12) /* Set cos(phi)=u, sin(phi)=sqrt(1-u^2), square both sides */

eqn4: 1-u^2= (0.7002075382097096*u + 0.2875852389075594)^2;

2 2

(%o12) 1 - u = (0.7002075382097096 u + 0.2875852389075594)

(%i13) expand(eqn4);

2 2

(%o13) 1 - u = 0.4902905965657019 u + 0.4027387043218267 u

+ 0.08270526963751801

(%i15) q1: rhs(%o13)-lhs(%o13);

2

(%o15) 1.490290596565702 u + 0.4027387043218267 u - 0.917294730362482

(%i17) soln5: solve(q1=0, u), numer;

(%o17) [u = - 0.9312186278281449, u = 0.6609769009797587]

(%i18) phi1: 180/%pi*acos(rhs(soln5[2])), numer;

(%o18) 48.62558056831977

Answer (a) (i) the plane to head at 48.6 deg

(north of east)

Extra Example 3.2A (4/4) (%i21) /* check answer */ phirad1: phi1*%pi/180, numer; (%o21) 0.8486764816109555 (%i22) vgx1: vgx, phi=phirad1, vw=230, vc=560; (%o22) 600.147064548665 (%i23) vgy1: vgy, phi=phirad1, vw=230, vc=560; (%o23) 420.2274986314038 (%i24) vg1: sqrt(vgx1^2+vgy1^2); (%o24) 732.6442859207912 (%i25) phi_g1: atan2(vgy1, vgx1)*180/%pi, numer; (%o25) 34.99999999999996

35 degrees is what we wanted!!!!! (%i26) t1: 3500/vg1; (%o26) 4.777215993162605

The trip takes 4.78 hours (typically 7.5 hours for New York London )