federica bianco
astro | data science | data for good
PHYS 207.013
Chapter 4
motion in a plane
Instructor: Dr. Bianco
TAs: Joey Betz; Lily Padlow
University of Delaware - Spring 2021
motion in 2D and 3D
Now that we are familiar with vectors, and equations of motion in 1D we can put them together and move in a 3D world (tho we will start with 2D)
H&R CH4 motion in 2D and 3D
There are no new concepts in this chapter: only application of established concepts in combination
So if something is not clear, go back to chapter 1, 2, and 3
position and displacement in 2D
H&R CH4 motion in 2D and 3D
Each position (at time t) can be described by a vector.
1) choose/identify the frame of reference (coordinates+origin)
2) a vector r starts at the origin and ends a the position
r_1
x
y
position and displacement in 2D
H&R CH4 motion in 2D and 3D
Each position (at time t) can be described by a vector.
1) choose/identify the frame of reference (coordinates+origin)
2) a vector r starts at the origin and ends a the position
r_1
x
y
position and displacement in 2D
H&R CH4 motion in 2D and 3D
Each position (at time t) can be described by a vector.
1) choose/identify the frame of reference (coordinates+origin)
2) a vector r starts at the origin and ends a the position
r_2
r_1
x
y
position and displacement in 2D
H&R CH4 motion in 2D and 3D
Each position (at time t) can be described by a vector.
1) choose/identify the frame of reference (coordinates+origin)
2) a vector r starts at the origin and ends a the position
r_1
r_2
x
y
position and displacement in 2D
H&R CH4 motion in 2D and 3D
Each position (at time t) can be described by a vector.
1) choose/identify the frame of reference (coordinates+origin)
2) a vector r starts at the origin and ends a the position
r_2
\Delta x
\Delta y
x
y
r_1
position and displacement in 2D
H&R CH4 motion in 2D and 3D
Each position (at time t) can be described by a vector.
1) choose/identify the frame of reference (coordinates+origin)
2) a vector r starts at the origin and ends a the position
\Delta x
\Delta y
\Delta \vec{r}
(x_1, y_1)
(x_2, y_2)
x
y
position and displacement in 2D
H&R CH4 motion in 2D and 3D
Each position (at time t) can be described by a vector.
1) choose/identify the frame of reference (coordinates+origin)
2) a vector r starts at the origin and ends a the position
\Delta \vec{r}
\Delta \vec{r} = \vec{r_2} - \vec{r_1} = (x_2-x_1)\hat{i} + (y_2-y_1)\hat{j}
x_1
x_2
y_2
y_1
(x_1, y_1)
(x_2, y_2)
position and displacement in 3D
H&R CH4 motion in 2D and 3D
Each position (at time t) can be described by a vector.
1) choose/identify the frame of reference (coordinates+origin)
2) a vector r starts at the origin and ends a the position
\Delta \vec{r}
(x_2, y_2, z_2)
\Delta \vec{r} = \vec{r_2} - \vec{r_1} = (x_2-x_1)\hat{i} + (y_2-y_1)\hat{j} + (z_2-z_1)\hat{k}
x_1
x_2
y_2
y_1
(x_1, y_1, z_1)
z_1
z_2
x
y
z
velocity in 3D
H&R CH4 motion in 2D and 3D
The velocity vector as well can be measured along each component
\Delta \vec{v} = \frac{\Delta(\vec{r})}{\Delta t} = \frac{(x_2-x_1)}{\Delta t}\hat{i} + \frac{(y_2-y_1)}{\Delta t}\hat{j} + \frac{(z_2-z_1)}{\Delta t}\hat{k}
\Delta \vec{r}
x_1
x_2
y_2
y_1
(x_1, y_1)
(x_2, y_2)
velocity in 3D
H&R CH4 motion in 2D and 3D
The velocity vector as well can be measured along each component
\Delta \vec{v} = \frac{\Delta(\vec{r})}{\Delta t} = \frac{\vec{r_2} - \vec{r_1}}{t_\mathrm{final} - t_\mathrm{initial}}
\Delta \vec{r}
x_1
x_2
y_2
y_1
(x_1, y_1)
(x_2, y_2)
H&R CH4 motion in 2D and 3D
The velocity vector as well can be measured along each component
\Delta \vec{v} = \frac{\Delta(\vec{r})}{\Delta t} = \frac{\Delta x}{\Delta t}\hat{i} + \frac{\Delta y}{\Delta t}\hat{j} + \frac{\Delta z}{\Delta t}\hat{k}
velocity in 3D
\Delta \vec{r}
x_1
x_2
y_2
y_1
(x_1, y_1)
(x_2, y_2)
H&R CH4 motion in 2D and 3D
The velocity vector as well can be measured along each component
\vec{v}(t=t) = \lim_{\Delta t \to 0} \frac{\Delta(\vec{r})}{\Delta t}
instantaneous velocity in 3D
\Delta \vec{r}
x_1
x_2
y_2
y_1
(x_1, y_1)
(x_2, y_2)
H&R CH4 motion in 2D and 3D
The velocity vector as well can be measured along each component
instantaneous velocity in 3D
\Delta \vec{r}
x_1
x_2
y_2
y_1
(x_1, y_1)
(x_2, y_2)
\vec{v}(t=t) = \lim_{\Delta t \to 0} \frac{\Delta(\vec{r})}{\Delta t} = \frac{d \vec{r}}{dt}
H&R CH4 motion in 2D and 3D
The velocity vector as well can be measured along each component
\Delta \vec{r}
\vec{v}(t=t) = \lim_{\Delta t \to 0} \frac{\Delta(\vec{r})}{\Delta t} = \frac{d \vec{r}}{d t} = \frac{d x}{d t}\hat{i} + \frac{d y}{d t}\hat{j} + \frac{d z}{d t}\hat{k}
x_1
x_2
y_2
y_1
instantaneous velocity in 3D
(x_1, y_1)
(x_2, y_2)
H&R CH4 motion in 2D and 3D
The velocity vector as well can be measured along each component
\Delta \vec{r}
\vec{v}(t=t) = \frac{d \vec{r}}{d t} = v_x\hat{i} + v_y\hat{j} + v_z\hat{k}
x_1
x_2
y_2
y_1
instantaneous velocity in 3D
(x_1, y_1)
(x_2, y_2)
MATH REVIEW:
derivatives
- the instantaneous velocity is described as the difference of vector (components) over an interval of time that "tends to zero" (is as small as possible)
- that is the derivative of the position with respect to time
MATH REVIEW:
derivatives
- the instantaneous velocity is described as the difference of vector (components) over an interval of time that "tends to zero" (is as small as possible)
- that is the derivative of the position with respect to time
v_\mathrm{av} = \frac{\Delta r}{\Delta t}
-> slope
if you need a review, this is pretty good! https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-1-new/ab-2-1/v/derivative-as-a-concept
\Delta{r}
\Delta{t}
t
r
MATH REVIEW:
derivatives
- the instantaneous velocity is described as the difference of vector (components) over an interval of time that "tends to zero" (is as small as possible)
- that is the derivative of the position with respect to time
v_\mathrm{av} = \frac{\Delta r}{\Delta t}
-> slope
if you need a review, this is pretty good! https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-1-new/ab-2-1/v/derivative-as-a-concept
\Delta{y}
\Delta{x}
x
y
MATH REVIEW:
derivatives
- the instantaneous velocity is described as the difference of vector (components) over an interval of time that "tends to zero" (is as small as possible)
- that is the derivative of the position with respect to time
v = \frac{d r}{d t}
-> slope of tangent
if you need a review, this is pretty good! https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-1-new/ab-2-1/v/derivative-as-a-concept
tangent
x
y
MATH REVIEW:
derivatives
- the instantaneous velocity is described as the difference of vector (components) over an interval of time that "tends to zero" (is as small as possible)
- that is the derivative of the position with respect to time
if you need a review, this is pretty good! https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-1-new/ab-2-1/v/derivative-as-a-concept
tangent
\Delta{y}
\Delta{x}
v = \frac{d r}{d t}
-> slope of tangent
x
y
MATH REVIEW:
derivatives
- the instantaneous velocity is described as the difference of vector (components) over an interval of time that "tends to zero" (is as small as possible)
- that is the derivative of the position with respect to time
if you need a review, this is pretty good! https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-1-new/ab-2-1/v/derivative-as-a-concept
tangent
\Delta{y}
\Delta{x}
v = \frac{d r}{d t}
-> slope of tangent
x
y
MATH REVIEW:
derivatives
derivative math rules https://www.rapidtables.com/math/calculus/derivative.html
\frac{d f(t)}{d t} = \frac{f(t + dt) - f(t)}{dt}
MATH REVIEW:
derivatives
derivative math rules https://www.rapidtables.com/math/calculus/derivative.html
\frac{d \vec{r}(t)}{d t} = \frac{\vec{r}(t + dt) - \vec{r}(t)}{dt}
MATH REVIEW:
derivatives
derivative math rules https://www.rapidtables.com/math/calculus/derivative.html
MATH REVIEW:
derivatives
derivative math rules https://www.rapidtables.com/math/calculus/derivative.html
\frac{d \vec{r}(t)}{d t} = \frac{\vec{r}(t + dt) - \vec{r}(t)}{dt}
\vec{r}(t) - \vec{r}_0 = \vec{v}(t) + \frac{1}{2} \vec{a}(t^2)
MATH REVIEW:
derivatives
derivative math rules https://www.rapidtables.com/math/calculus/derivative.html
\frac{d c}{d t} = 0
\frac{d \vec{r}(t)}{d t} = \frac{\vec{r}(t + dt) - \vec{r}(t)}{dt}
\frac{d c f(t)}{d t} = c\frac{d f(t)}{d t}
MATH REVIEW:
derivatives
derivative math rules https://www.rapidtables.com/math/calculus/derivative.html
f(t) = t^c =>\frac{d f(t)}{d t} = c t^{c-1}
\frac{d \vec{r}(t)}{d t} = \frac{\vec{r}(t + dt) - \vec{r}(t)}{dt}
\frac{d t^2}{d t} = 2 t^{2-1}
\frac{d t}{d t} = 1
MATH REVIEW:
derivatives
derivative math rules https://www.rapidtables.com/math/calculus/derivative.html
r = ct => v = \frac{d(c\cdot t)}{d t} = c
a = \frac{d^2 (c\cdot t^2)}{d t^2} = \frac{d(2 c\cdot t)}{dt} = 2c
v = \frac{d (c\cdot t^2)}{d t} = 2 c\cdot t
r = c t^2 =>
{
\frac{d \vec{r}(t)}{d t} = \frac{\vec{r}(t + dt) - \vec{r}(t)}{dt}
\vec{r}(t) - \vec{r}_0 = \vec{v}(t) + \frac{1}{2} \vec{a}(t^2)
f(t) = t^c =>\frac{d f(t)}{d t} = c t^{c-1}
\frac{d c}{d t} = 0
\frac{d c f(t)}{d t} = c\frac{d f(t)}{d t}
MATH REVIEW:
derivatives
derivative math rules https://www.rapidtables.com/math/calculus/derivative.html
\frac{d \sin{t}}{d t} = \cos{t}
v = \frac{d(c\sin{t})}{d t} = c \cos{t}
\frac{d \cos{t}}{d t} = -\sin{t}
a = \frac{d^2(c\sin{t})}{d t^2} = \frac{d(c\cos{t})}{d t} = -c \sin{t}
r = c \sin{t}
{
\frac{d \vec{r}(t)}{d t} = \frac{\vec{r}(t + dt) - \vec{r}(t)}{dt}
H&R CH4 motion in 2D and 3D
example
\vec{r}(t=2s)
\vec{r}(0)
2
6
x
y
z
-16
H&R CH4 motion in 2D and 3D
example
I can already tell:
- the x components is linear in time: =>constant velocity along the x axis
- the y component has a factor t^2: =>accelerated motion along the y axis
- the z components has no factors of t: =>there is no motion along the z axis
\vec{r}(t=2s)
\vec{r}(0)
2
6
x
y
z
-16
example
x
H&R CH4 motion in 2D and 3D
\vec{r}(t=2s)
\vec{r}(0)
2
6
x
y
z
-16
example
x
y
H&R CH4 motion in 2D and 3D
\frac{d(ct)}{dt} = c
\frac{d(ct^a)}{dt} = c a t^{a-1}
\frac{d(ct)}{dt} = c
\frac{d(c)}{dt} = 0
math relations and derivatives
\vec{r}(t=2s)
\vec{r}(0)
2
6
x
y
z
-16
example
x
z
H&R CH4 motion in 2D and 3D
-16
\theta = \arctan {\frac{v_y}{v_x}} = \arctan {\frac{v \sin\theta}{v \cos\theta}}
v = \sqrt{v_x^2 + v_y^2 + v_z^2}
v_x = v \cos(\theta) ; \\
v_y = v \sin{\theta}
geometric relations and trig
\theta = \arctan {\frac{v_y}{v_x}} = \arctan {\frac{v \sin\theta}{v \cos\theta}}
\vec{r}(t=2s)
\vec{r}(0)
2
6
x
y
z
-16
KEY POINTS:
- the average velocity is described as the difference of vector (components) over time
- the instantaneous velocity is described as the difference of vector (components) over an interval of time that "tends to zero" (is as small as possible)
- that is the derivative of the position with respect to time
- position is described by vectors
- displacement on each axis is described as the difference of vector components
motion in 2D and 3D
H&R CH4 motion in 2D and 3D
Projectile motion
\theta
\theta
r
SPECIAL CASES OF MOTION IN 2D
Uniform cirular motion
Projectile motion
\theta
\vec{v_0} = v_0 \cos{\theta} \hat{i} + v_0 \sin{\theta} \hat{j}
SPECIAL CASES OF MOTION IN 2D
H&R CH4 motion in 2D and 3D
\theta
r
Uniform cirular motion
Projectile motion
\theta
\vec{v_0} = v_0 \cos{\theta} \hat{i} + v_0 \sin{\theta} \hat{j}
\vec{a} = -9.8 \hat{j} \frac{m}{s}
SPECIAL CASES OF MOTION IN 2D
H&R CH4 motion in 2D and 3D
\theta
r
Uniform cirular motion
Projectile motion
\theta
\vec{v_0} = v_0 \cos{\theta} \hat{i} + v_0 \sin{\theta} \hat{j}
\vec{v} = -v \sin{\theta} \hat{i} + v \cos{\theta} \hat{j}
SPECIAL CASES OF MOTION IN 2D
H&R CH4 motion in 2D and 3D
\theta
r
Uniform cirular motion
\vec{a} = -9.8 \hat{j} \frac{m}{s}
Projectile motion
\theta
\vec{v_0} = v_0 \cos{\theta} \hat{i} + v_0 \sin{\theta} \hat{j}
\vec{a} = \left(-\frac{v^2}{r} \cos{\theta}\right) \hat{i} + \left(-\frac{v^2}{r} \sin{\theta}\right)\hat{j}
\vec{v} = -v \sin{\theta} \hat{i} + v \cos{\theta} \hat{j}
SPECIAL CASES OF MOTION IN 2D
H&R CH4 motion in 2D and 3D
\theta
r
Uniform cirular motion
\vec{a} = -9.8 \hat{j} \frac{m}{s}
PROJECTILE MOTION
\theta
\vec{v_0} = v_0 \cos{\theta} \hat{i} + v_0 \sin{\theta} \hat{j}
\vec{a} = - 9.8 \hat{j} \frac{m}{s^2}
Think about this the same way we thought of free-fall motion in Chap2: it is just regular motion, for which I have some extra info: that v is constant and a is only on the y axis a=g
- constant motion along x
- accelerate (downward) motion along y
H&R CH4 motion in 2D and 3D
SPECIAL CASES OF MOTION IN 2D
Projectile motion
\theta
\vec{v_0} = v_0 \cos{\theta} \hat{i} + v_0 \sin{\theta} \hat{j}
Toni Stone (July 17, 1921 – November 2, 1996) was the first woman to play as a regular on an American big-league professional baseball team
they
\vec{a} = - 9.8 \hat{j} \frac{m}{s^2}
PROJECTILE MOTION
\theta
\vec{v_0} = v_0 \cos{\theta} \hat{i} + v_0 \sin{\theta} \hat{j}
R = \frac{v_0^2}{g} \sin{2\theta_0}
range motion tells me the distance traveled horizontally
**if the motion starts and ends at y=0**
R
(note: use the trig identity )
2\sin{\theta}\cos{\theta} = \sin{2\theta}
H&R CH4 motion in 2D and 3D
\vec{a} = - 9.8 \hat{j} \frac{m}{s^2}
PROJECTILE MOTION
\vec{v_0} = v_0 \cos{\theta} \hat{i} + v_0 \sin{\theta} \hat{j}
R
H&R CH4 motion in 2D and 3D
when =>
y = y_{\mathrm{max}}
v_y(y = y_\mathrm{max}) = 0
\theta = 0\\
v_y = v\sin{\theta}
\sin{0} = 0
}
math:
because v changes direction => changes sign => has to go through 0
physics:
\vec{a} = - 9.8 \hat{j} \frac{m}{s^2}
PROJECTILE MOTION
\theta
\vec{v_0} = v_0 \cos{\theta} \hat{i} + v_0 \sin{\theta} \hat{j}
R
H&R CH4 motion in 2D and 3D
I recommend you try solving problems like this one, where the initial motion has no upward velocity - these problems tend to be more confusing perhaps
\vec{a} = - 9.8 \hat{j} \frac{m}{s^2}
KEY POINTS:
- the velocity along y is zero at max height
- projectile motion is motion under some initial velocity in 2D and downward acceleration along the y axis (no acceleration in x)
- I can calculate how far in x a projectile starting from y=0 would land (i.e. second y=0 solution) if I know the initial velocity and angle at which it was thrown (range)
H&R CH4 motion in 2D and 3D
R = \frac{v_0^2}{g} \sin{2\theta_0}
v_x = \mathrm{const}: x - x0 = v_{0,x} t\\
v_y = \mathrm{accelerated}: y - y0 = v_{0,y} t - \frac{1}{2}a t^2
parabolic motion
UNIFORM CIRCULAR MOTION
\vec{a} = \left(-\frac{v^2}{r} \cos{\theta}\right) \hat{i} + \left(-\frac{v^2}{r} \sin{\theta}\right)\hat{j}
\vec{v} = -v \sin{\theta} \hat{i} + v \cos{\theta} \hat{j}
H&R CH4 motion in 2D and 3D
\theta
r
Uniform cirular motion
SPEED DOES NOT CHANGE - VELOCITY CHANGES!
UNIFORM CIRCULAR MOTION
\vec{v} = -v \sin{\theta} \hat{i} + v \cos{\theta} \hat{j}
H&R CH4 motion in 2D and 3D
r
Uniform cirular motion
T period - how long it takes to go around once [T] (sec)
F frequency - how many revolutions per unit time [1/T] (1/sec or Hz)
F = 1/T
SPEED DOES NOT CHANGE - VELOCITY CHANGES!
}
\theta
UNIFORM CIRCULAR MOTION
\vec{v} = -v \sin{\theta} \hat{i} + v \cos{\theta} \hat{j}
H&R CH4 motion in 2D and 3D
r
T period - how long it takes to go around once [T] (sec)
F frequency - how many revolutions per unit time [1/T] (1/sec or Hz)
F = 1/T
angular velocity: rate of change of the angle : [angle/T] (randians / s)
\omega = \frac{2\pi}{T}
vc = circumference / period = [L/T] (m/s)
}
\omega = \frac{2\pi}{T}
v_c = \frac{2\pi r}{T}
}
\omega = \frac{2\pi}{T}
v_c = \frac{2\pi r}{T} = \omega r
\alpha
v_c = \frac{2\pi r}{T}
Uniform cirular motion
SPEED DOES NOT CHANGE - VELOCITY CHANGES!
\alpha
UNIFORM CIRCULAR MOTION
\vec{v} = -v \sin{\theta} \hat{i} + v \cos{\theta} \hat{j}
H&R CH4 motion in 2D and 3D
r
Uniform cirular motion
SPEED DOES NOT CHANGE - VELOCITY CHANGES!
\alpha
UNIFORM CIRCULAR MOTION
\vec{v} = -v \sin{\theta} \hat{i} + v \cos{\theta} \hat{j}
H&R CH4 motion in 2D and 3D
r
Uniform cirular motion
SPEED DOES NOT CHANGE - VELOCITY CHANGES!
\alpha
x(t) = r \cos{\omega t}
y(t) = r \sin{\omega t}
simple harmonic motion
UNIFORM CIRCULAR MOTION
\vec{a} = \left(-\frac{v^2}{r} \cos{\theta}\right) \hat{i} + \left(-\frac{v^2}{r} \sin{\theta}\right)\hat{j}
\vec{v} = -v \sin{\theta} \hat{i} + v \cos{\theta} \hat{j}
H&R CH4 motion in 2D and 3D
r
Uniform cirular motion
SPEED DOES NOT CHANGE - VELOCITY CHANGES!
acceleration:
|a| = \frac{|v_c|^2}{r}
\theta
angular velocity: rate of change of the angle : [angle/T] (randians / s)
\omega = \frac{2\pi}{T}
\alpha
vc = circumference / period = [L/T] (m/s)
v_c = \frac{2\pi r}{T}
\omega = \frac{2\pi}{T}
v_c = \frac{2\pi r}{T}
}
\omega = \frac{2\pi}{T}
v_c = \frac{2\pi r}{T} = \omega r
UNIFORM CIRCULAR MOTION
\vec{a} = \left(-\frac{v^2}{r} \cos{\theta}\right) \hat{i} + \left(-\frac{v^2}{r} \sin{\theta}\right)\hat{j}
\vec{v} = -v \sin{\theta} \hat{i} + v \cos{\theta} \hat{j}
H&R CH4 motion in 2D and 3D
r
Uniform cirular motion
SPEED DOES NOT CHANGE - VELOCITY CHANGES!
\theta
what is the relationship between a and the radius?
acceleration:
\omega = \frac{2\pi}{T}
v_c = \frac{2\pi r}{T}
}
\omega = \frac{2\pi}{T}
v_c = \frac{2\pi r}{T} = \omega r
|a| = \frac{|v_c|^2}{r}
|a| \propto \frac{1}{r} |v_c|^2 = \frac{1}{r}(\omega r)^2 = \omega^2 r \\
|a| \propto r
UNIFORM CIRCULAR MOTION
\vec{a} = \left(-\frac{v^2}{r} \cos{\theta}\right) \hat{i} + \left(-\frac{v^2}{r} \sin{\theta}\right)\hat{j}
\vec{v} = -v \sin{\theta} \hat{i} + v \cos{\theta} \hat{j}
H&R CH4 motion in 2D and 3D
r
Uniform cirular motion
angular velocity: rate of change of the angle: [angle/T] (randians / s)
\omega = \frac{2\pi}{T}
v = circumference / period = [L/T] (m/s)
v = \frac{2\pi r}{T}
SPEED DOES NOT CHANGE - VELOCITY CHANGES!
LOOK AT THE DERIVATION OF a FROM v IN THE BOOK!
\theta
r
H&R CH4 motion in 2D and 3D
acceleration:
|a| = \frac{|v_c|^2}{r}
\omega = \frac{2\pi}{T}
v_c = \frac{2\pi r}{T}
}
\omega = \frac{2\pi}{T}
v_c = \frac{2\pi r}{T} = \omega r
UNIFORM CIRCULAR MOTION
\vec{a} = \left(-\frac{v^2}{r} \cos{\theta}\right) \hat{i} + \left(-\frac{v^2}{r} \sin{\theta}\right)\hat{j}
\vec{v} = -v \sin{\theta} \hat{i} + v \cos{\theta} \hat{j}
r
r
90 deg angles:
angles:
\theta
\theta
\theta
\theta
r
UNIFORM CIRCULAR MOTION
\vec{a} = \left(-\frac{v^2}{r} \cos{\theta}\right) \hat{i} + \left(-\frac{v^2}{r} \sin{\theta}\right)\hat{j}
\vec{v} = -v \sin{\theta} \hat{i} + v \cos{\theta} \hat{j}
Uniform cirular motion
r
\theta
\theta
\vec{a} = a \cos\theta \hat{i} + a \sin \theta \hat{j}
\theta
r
a
UNIFORM CIRCULAR MOTION
\vec{a} = \left(-\frac{v^2}{r} \cos{\theta}\right) \hat{i} + \left(-\frac{v^2}{r} \sin{\theta}\right)\hat{j}
\vec{v} = -v \sin{\theta} \hat{i} + v \cos{\theta} \hat{j}
H&R CH4 motion in 2D and 3D
r
Uniform cirular motion
at what x coordinate will the ball fall?
0
r/2
r
3r/2
-r/2
-r
-3r/2
-2r
2r
>2r
<-2r
\theta
UNIFORM CIRCULAR MOTION
\vec{a} = \left(-\frac{v^2}{r} \cos{\theta}\right) \hat{i} + \left(-\frac{v^2}{r} \sin{\theta}\right)\hat{j}
\vec{v} = -v \sin{\theta} \hat{i} + v \cos{\theta} \hat{j}
H&R CH4 motion in 2D and 3D
r
Uniform cirular motion
0
r/2
r
3r/2
-r/2
-r
-3r/2
-2r
2r
>2r
<-2r
\theta
Q: Where on the x axis would a particle released from circular motion at the position indicated in the diagram to the right
A: The velocity vector v at the position of release is perpendicular to the position vector r . Thus after release this problem becomes a projectile motion problem with
v_0 = v_y \hat{j}; v_x = 0
Since there is no x displacement and the x position is x=r for the entire motion
v_0 = v_y \hat{j}
UNIFORM CIRCULAR MOTION
\vec{a} = \left(-\frac{v^2}{r} \cos{\theta}\right) \hat{i} + \left(-\frac{v^2}{r} \sin{\theta}\right)\hat{j}
\vec{v} = -v \sin{\theta} \hat{i} + v \cos{\theta} \hat{j}
H&R CH4 motion in 2D and 3D
r
Uniform cirular motion
\theta
UNIFORM CIRCULAR MOTION
\vec{a} = \left(-\frac{v^2}{r} \cos{\theta}\right) \hat{i} + \left(-\frac{v^2}{r} \sin{\theta}\right)\hat{j}
H&R CH4 motion in 2D and 3D
\theta
r
Uniform cirular motion
1. A centripetal acceleration pointing toward the center of the circular orbit keeps the body moving at constant speed
acceleration:
|a| = \frac{|v_c|^2}{r}
UNIFORM CIRCULAR MOTION
\vec{a} = \left(-\frac{v^2}{r} \cos{\theta}\right) \hat{i} + \left(-\frac{v^2}{r} \sin{\theta}\right)\hat{j}
H&R CH4 motion in 2D and 3D
\theta
r
Uniform cirular motion
1. A centripetal acceleration pointing toward the center of the circular orbit keeps the body moving at constant speed
acceleration:
|a| = \frac{|v_c|^2}{r}
2. By Newton's II law: F = ma if there is an acceleration there is a force!
Force:
F = m|a| = m\frac{|v_c|^2}{r}
UNIFORM CIRCULAR MOTION
\vec{a} = \left(-\frac{v^2}{r} \cos{\theta}\right) \hat{i} + \left(-\frac{v^2}{r} \sin{\theta}\right)\hat{j}
H&R CH4 motion in 2D and 3D
\theta
r
Uniform cirular motion
1. A centripetal acceleration pointing toward the center of the circular orbit keeps the body moving at constant speed
acceleration:
|a| = \frac{|v_c|^2}{r}
2. By Newton's II law: F = ma if there is an acceleration there is a force!
Force:
F = m|a| = m\frac{|v_c|^2}{r}
F \propto \frac{1}{r}
UNIFORM CIRCULAR MOTION
\vec{a} = \left(-\frac{v^2}{r} \cos{\theta}\right) \hat{i} + \left(-\frac{v^2}{r} \sin{\theta}\right)\hat{j}
H&R CH4 motion in 2D and 3D
\theta
r
Uniform cirular motion
1. A centripetal acceleration pointing toward the center of the circular orbit keeps the body moving at constant speed
acceleration:
|a| = \frac{|v_c|^2}{r}
2. By Newton's II law: F = ma if there is an acceleration there is a force!
Force:
F = m|a| = m\frac{|v_c|^2}{r}
F \propto \frac{1}{r}
UNIFORM CIRCULAR MOTION
r
F = m|a| = m\frac{|v_c|^2}{r}
F \propto \frac{1}{r}
KEY POINTS:
- the acceleration is toward the center of the circle and constant in magnitude
- uniform circular motion is the motion of an object with velocity v constant in magnitude and uniformly changing in direction
- we define the period T frequency F (inverse of the period) and angular velocity omega
H&R CH4 motion in 2D and 3D
v = \frac{2\pi r}{T} = \omega r
\vec{v} = -v \sin{\theta} \hat{i} + v \cos{\theta} \hat{j}
\vec{a} = \left(-\frac{v^2}{r} \cos{\theta}\right) \hat{i} + \left(-\frac{v^2}{r} \sin{\theta}\right)\hat{j}
uniform circular motion
- the velocity vector always points tangentially to the circle
KEY POINTS:
- the relative motion of two objects is given by the sum of the motions along each axis
H&R CH4 motion in 2D and 3D
relative motion
MATH REVIEW:
integrals
- the inverse operation to the derivative is the integral
- The integral of a curve is calculated as the difference of the integral at the two extreme points
\int_a^b f^{'}(x)dx = f(b) - f(a)
MATH REVIEW:
integrals
based on the rules we learned for derivatives we can figure out what the value of the integral is for specific functions
\frac{d f(t) + g(t)}{d t} = \frac{d f(t)}{d t} + \frac{d g(t)}{d t}
\frac{d t}{d t} = 1
\frac{d t^2}{d t} = 2 t
t + \mathrm{const}
\frac{1}{2}t^2 + \mathrm{const}
\frac{d c}{d t} = 0
\int{1}{dt} =
\frac{d t^c}{d t} = c t^{c-1}
\int t {dt} =
MATH REVIEW:
integrals
based on the rules we learned for derivatives we can figure out what the value of the integral is for specific functions
\frac{d f(t) + g(t)}{d t} = \frac{d f(t)}{d t} + \frac{d g(t)}{d t}
\frac{d t}{d t} = 1
\frac{d t^2}{d t} = 2 t
t + \mathrm{const} = t(a)-t(b) + const
\frac{1}{2}t^2 + \mathrm{const} = t^2(a) - t^2(b) + const
\frac{d c}{d t} = 0
\int{1}{dt} =
\frac{d t^c}{d t} = c t^{c-1}
\int t {dt} =
MATH REVIEW:
integrals
based on the rules we learned for derivatives we can figure out what the value of the integral is for specific functions
\frac{d f(t) + g(t)}{d t} = \frac{d f(t)}{d t} + \frac{d g(t)}{d t}
\frac{d t}{d t} = 1
\frac{d t^2}{d t} = 2 t
\int t ^2{dt} = \frac{1}{3}t^3 + \mathrm{const}
\frac{d c}{d t} = 0
\frac{d t^c}{d t} = c t^{c-1}
\int t^2 {dt} =
\int t ^c{dt} = \frac{1}{c}t^c + \mathrm{const}
MATH REVIEW:
integrals
\frac{d \sin{t}}{d t} = \cos{t}
\frac{d \cos{t}}{d t} = -\sin{t}
MATH REVIEW:
integrals
based on the rules we learned for derivatives we can figure out what the value of the integral is for specific functions
\int sin(t) =
sin(t) + const
\int cos(t) =
-cos(t) + const
MATH REVIEW:
integrals
MATH REVIEW:
integrals
based on the rules we learned for derivatives we can figure out what the value of the integral is for specific functions
\vec{v} = \frac{d(\vec{x})}{dt}
\vec{a} = \frac{d(\vec{v})}{dt}
\vec{v} = \int{\vec{a}}{dt}
\vec{x} = \int{\vec{v}}{dt}
phys207 2D-3D motion
By federica bianco
phys207 2D-3D motion
vectors, vector math, trigonometry
