��� If your clicker answers are not being recorded on Blackboard, let us

```•  If your clicker answers are not being recorded on Blackboard, let us know. •  There is a set of prac;ce problems on vectors in MasteringPhysics. There are also some more in online HW1 (s;ll accessible if you haven’t done them). •  Please make sure your Phys 221 Student ID in MP is your ISU netID, not your ISU email or ISU student ID #. (This is diﬀerent than your MP login ID.) You can change it in your Proﬁle. Example: Maximum height A person throws a ball from a window 10 m above the ground, with an ini9al velocity of 10 m/s and at a 30-­‐degree angle with the horizontal. Air resistance can be neglected. Calculate the maximum height the ball will reach above the ground. v0 = 10 m/s ymax ⇔ v y = 0
y y0 = 10 m θ = 300 Time to get to the top:
0 = v0 sinθ − gt → t =
ymax ? v0 sinθ
g
x 1
ymax = y0 + v0 sinθ t − gt 2
2
2
What determines the maximum height of a projec9le? Answer: the ini9al velocity v 0y in the ver9cal direc9on v sinθ 1 ⎛ v0 sinθ ⎞
= y0 + v0 sinθ 0
− g⎜
g
2 ⎝ g ⎟⎠
2
(10 m/s ) sin2 (30° )
v02 sin2 θ
= y0 +
= 10 m +
= 11.3 m
2
2g
2 ( 9.8 m/s )
( )
How does the range R depend on θ ? y v0 yﬁnal = 0 θ 1
y = y0 + v0 y t + ay t 2
2
1
0 = 0 + v0 y t − gt 2
2
x = x0 + v 0 x t
x ⎧t = 0 (start!)
Transfer ;me (t) from ver;cal mo;on ⎪
⎨ 2v0 y 2v0 sinθ
to horizontal mo;on ⎪t = g =
g
⎩
R = x − x0 = v 0 x t
2v0 sinθ 2v02 sinθ cosθ v 02 sin2θ
= v0 cosθ
=
=
g
g
g
R 450 900 Angle (θ) Maximum range is for θ = 450 Lecture -­‐ 06 Circular mo9on Review: accelera;on in 2D (and 3D) !
!
! dv d(vv)
ˆ
dv
dvˆ
a=
=
=
vˆ + v
v = vvˆ
dt
dt
dt
dt
! ! !
a = a|| + a⊥
a|| or tangen9al accelera9on: Parallel/an9parallel to velocity Changes the speed. aperp or normal or centripetal/
radial accelera9on: •  Perpendicular to velocity •  Inwards (towards the center of curvature) •  Does not change the speed but shiVs the direc9on of the mo9on. Circular mo;on Circular mo9on is the mo9on in a circle with constant radius. Polar coordinates Polar coordinates (r, θ ) are more convenient than Cartesian coordinates to describe circular mo;on: r = R, only θ = θ (t) Arch: s = R θ this equa;on only works if θ is in radians Deﬁni;on: y
•  1 radian = angle so that s=R •  1 revolu;on = 2π radians Rela9on to Cartesian coordinates: x = r cos θ y = r sin θ R
θ s
x
Velocity dx
dy
Cartesian coordinates: v x = ; vy = dt
dt
Polar coordinates: dr
dθ
vr = ; ω =
dt
dt
Radial velocity Angular velocity For circular mo9on: vr = 0 ds
dθ
s = Rθ ⎯⎯
→ =R
⎯⎯
→ v = Rω
dt
dt
(where ω is in radians/unit 9me) EXAMPLE: Two balls Two balls connected by thin rod as shown, at distances R and 2R from the center, move in circles. Same angular speed ω for both (same angle in any Δt) Diﬀerent (linear) speeds (ball 2 travels twice the distance in any Δt) v1 = Rω
v2 = 2Rω
Uniform circular mo;on (UCM) ω is constant. This is an example of periodic mo;on.
Period T : Time it takes to go back to the same situa9on (same posi9on, same velocity). Frequency f : Number of revolu9ons per unit 9me. Units: Hz (turns per second), rpm (rev per min) 2π
T=
ω
1
f=
T
ω = 2π f
Example: Ferris Wheel A Ferris Wheel of radius 8.0 m rotates at a constant rate of 1.5 rpm. Find: a. The period b. The linear speed of a cabin. ACT: Ferris wheel The ferris wheel in the ﬁgure rotates counterclockwise at a uniform rate. What is the direc;on of the average accelera;on of a gondola as it goes from the top to the bo`om of its trajectory? A. Down B. C. The accelera9on is 0 because the mo9on is uniform. Radial or centripetal accelera;on During uniform circular mo9on, speed is constant, but velocity is not!!! The direc9on keeps changing!   
a =ûa || + a⊥
centripetal accelera9on •  Points to the center of the circle UCM in Cartesian components Angle covered in 9me interval ∆t: θ= θ0 + ω ∆t. If we choose θ0 = 0 at t = 0, it’s θ = ω t y x ωt y
t x The Cartesian coordinates are sine-­‐func9ons: ( )
y = Rsin(ω t )
x = Rcos ω t
Magnitude of Radial Accelera;on For UCM: x = R cos (ωt )
y = R sin (ωt )
ax = −Rω 2 cos (ωt )
v x = −Rω sin (ωt )
d
dt
v y = Rω cos (ωt )
|r |= R
d
dt
| a |= Rω 2
!
!
a = −ω 2 r = −ω 2 R rˆ
|v |= Rω
ay = −Rω 2 sin (ωt )
| a |= Rω 2


a = −ω 2 r = −ω 2 R rˆ
| a |= Rω 2


a = −ω 2 r = −ω 2 R rˆ
|a|= Rω
In UCM, all the accelera9on is centripetal. Thus, 2
v
ar = Rω 2 =
R
Accelera;on simulator In the movie “The Right Stuﬀ”, a seat at the end of a long arm that rotates very fast is used to prepare astronauts for high accelera9ons. If R = 5 m, what is the speed needed to have a = 5g? ω R ACT: Angular speed A horizontal rod of length 1 m turns with constant speed about a ver9cal axis. The blue 9p of the rod makes one turn each second. Find the angular speed of the red spot painted in the middle of the rod. A.  π rad/s B.  2π rad/s C.  4π rad/s The linear speed is not: vblue tip = R ω = (1 m)(2π rad/s) = 6.28 m/s
vred spot = R ω = (0.5 m)(2π rad/s) = 3.14 m/s
6.28 m
3.14 m
1 m
It makes sense: The blue 9p is covering twice the distance covered by the red spot in the same 9me. v2
Does this make sense? |ar |=
R
1
As you decrease the radius, the rate at |a|∝
which the velocity shiVs, hence the R
accelera9on, grows. As you increase the velocity, the |a|∝ v 2
accelera9on grows in two ways: -­‐ The rate at which the velocity is shiVing grows. -­‐ The amount of velocity which needs to be shiVed around also grows. R Non-­‐uniform circular mo;on Slowing down R
ar
v
Speeding up v
a a at
at
R
r
a
Radial accelera9on (towards the center) → changes in direc9on Tangen9al accelera9on (tangen9al to trajectory) → changes in speed Angular accelera;on If linear speed v is changing, the angular speed ω is also changing Angular accelera9on dω d 2θ
α=
= 2
dt dt
d
dt
v = Rω ⎯⎯
→ at = Rα
!
! dv
dv
because'at = '''''''(not'to'be'confused'with'a =
''!!!)
dt
dt
Equa;ons of mo;on when α= constant The deriva9on is formally iden9cal to what we did in 1D: dθ
ω=
dt
dω
α=
dt
ω = ω0 + α t
1 2
θ = θ 0 + ω0 t + α t
2
Δθ ω + ω 0
ω=
=
Δt
2
ω 2 − ω02 = 2αΔθ
Example: Wheel At t = 0, a wheel has an angular velocity of 20 rad/s. It speeds up at a uniform rate of 4.0 rad/s2 un9l a circuit breaker trips at t = 5.0 s. From then on, it turns through 1600 rad as it coasts to a stop with constant angular accelera9on. What was its accelera9on as it slowed down? Example Test Ques;on(ETQ) ```