PHYS 207.013

Chapter 10

rotation

Instructor: Dr. Bianco

TAs: Joey Betz; Lily Padlow

University of Delaware - Spring 2021

https://slides.com/federicabianco/phys20k713_10

Equity in physics classes

You have the right to the space that you need to succeed based on your identity

Equity in physics classes

How is it possible that the canon of thought in all the disciplines of the Social Sciences and Humanities in the Westernized university [...] is based on the knowledge produced by a few men from five countries in Western Europe (Italy, France, England, Germany and the USA)? How is it possible that men from these five countires achieved such an epistemic privilege to the point that their knowledge today is considered superior over the knowledge of the rest of the world? How did they come to monopolize the authority of knowledge in the world? Why is it that what we know today as social, historical, philosophical, or Critical Theory is based on the socio-historical experience and world views of men from these five countries?

Epistemic Racism/Sexism, Westernized Universities and the Four Genocides/Epistemicides of the Long 16th Century

Ramón Grosfoguel

Equity in physics classes

Decolonizing science

Think about examples of knowledge related to physics that do not come from European-Western culture

ZOOM breakout rooms of 4-5 students

Decolonizing science

Think about examples of knowledge related to physics that do not come from European-Western culture

the invention of 0 is indian - also mayan arab and chinese - decidedly NOT European

Decolonizing science

Think about examples of knowledge related to physics that do not come from European-Western culture

In fact while Arabs codified and diffused algebra the Europe was in the "Dark Ages"

Decolonizing science

Think about examples of knowledge related to physics that do not come from European-Western culture

Nilakantha Somayaji 14 June 1444 – 1544

discovers an infinite series for π.

conceptualizes heliocentric system

Decolonizing science

Think about examples of knowledge related to physics that do not come from European-Western culture

12th century AD: Jewish polymath Baruch ben Malka, later Abu'l-Barakāt al-Baghdādī, in Iraq formulates a qualitative form of Newton's second law for constant forces.

An explanation of the acceleration of falling bodies as the accumulation of successive increments of power (ref) === Newton laws 300 years early!

Decolonizing science

Think about examples of knowledge related to physics that do not come from European-Western culture

Until recently, people believed that vulcanization–combining rubber with other materials to make it more durable–was discovered by the American Charles Goodyear in the 19th century. However, historians now think that the Maya were producing rubber products about 3,000 years before Goodyear received his patent in 1843. https://www.history.com/topics/ancient-americas/mayan-scientific-achievements

Decolonizing science

Think about examples of knowledge related to physics that do not come from European-Western culture

https://twitter.com/AboriginalAstro

https://beatricemartini.it/blog/decolonizing-technology-reading-list/

https://en.wikipedia.org/wiki/Timeline_of_scientific_discoveries

https://forms.gle/k8aDPG5MygDdch3g7

HOMEWORK

Equity in physics classes

You have the right to the space that you need to succeed based on your identity

https://engr.udel.edu/about/diversity-inclusion/

Equity in physics classes

You have the right to the space that you need to succeed based on your identity

https://dpacdeicommittee.github.io/

rotation

H&R CH10 rotation

\mathrm{angular~ position} ~~\theta = \frac{s}{r} \frac{[L]}{[L]}

radians

radians

2 \pi = 360^o

degree

=> radians are revolutions

rotation

H&R CH10 rotation

\mathrm{angular~ position} ~~\theta = \frac{s}{r} \frac{[L]}{[L]}

radians

2 \pi = 360^o

angular displacement:

\theta_1

\theta_2

\theta_2 -\theta_1

rotation

H&R CH10 rotation

\mathrm{angular~ position} ~~\theta = \frac{s}{r} \frac{[L]}{[L]}

radians

2 \pi = 360^o

\theta_1

\theta_2

angular displacement:

\theta_2 -\theta_1

angular velocity:

\bar{\omega} = \frac{\theta_2 -\theta_1}{t_2 - t_1} = \frac{\Delta \theta}{\Delta t}

rotation

H&R CH10 rotation

\mathrm{angular~ position} ~~\theta = \frac{s}{r} \frac{[L]}{[L]}

radians

2 \pi = 360^o

\theta_1

\theta_2

\omega = lim_{\Delta t -> 0} \frac{\Delta \theta}{\Delta t} = \frac{d\theta}{dt}

angular displacement:

\theta_2 -\theta_1

angular velocity:

\bar{\omega} = \frac{\theta_2 -\theta_1}{t_2-t_1} = \frac{\Delta \theta}{\Delta t}

rotation

H&R CH10 rotation

\mathrm{angular~ position} ~~\theta = \frac{s}{r} \frac{[L]}{[L]}

radians

2 \pi = 360^o

\theta_1

\theta_2

\alpha = lim_{\Delta t -> 0} \frac{\Delta \omega}{\Delta t} = \frac{d\omega}{dt}

angular velocity:

\omega

angular acceleration:

\bar{\alpha} = \frac{\omega -\omega}{t_2 - t_1} = \frac{\Delta \omega}{\Delta t}

rotation

H&R CH10 rotation

\mathrm{angular~ position} ~~\theta = \frac{s}{r} \frac{[L]}{[L]}

radians

radians

2 \pi = 360^o

degree

circumference:

2 \pi r

rotation

H&R CH10 rotation

\mathrm{angular~ position} ~~\theta = \frac{s}{r} \frac{[L]}{[L]}

radians

radians

2 \pi = 360^o

degree

circumference:

2 \pi r

linear speed:

v_l = \frac{2 \pi r}{\Delta t}

MATH REVIEW:

trigonometry

vector component: the eq.s of motion you studied in chapter 2 have a parallel here

kinematic of circular motion

H&R CH10 rotation

\mathrm{angular~ position} ~~\theta = \frac{s}{r} \frac{[L]}{[L]}

radians

s = \theta r

\frac{ds}{dt} = \frac{\theta r}{dt} = \frac{d \theta}{dt} r

v~~~~~~=

\omega r

H&R CH10 rotation

\mathrm{angular~ position} ~~\theta = \frac{s}{r} \frac{[L]}{[L]}

radians

s = \theta r

\frac{ds}{dt} = \frac{\theta r}{dt} = \frac{d \theta}{dt} r

v~~~~~~=

\omega r

T = \frac{2\pi r}{v} = \frac{2 \pi}{\omega}

kinematic of circular motion

H&R CH10 rotation

\mathrm{angular~ position} ~~\theta = \frac{s}{r} \frac{[L]}{[L]}

radians

s = \theta r

\frac{ds}{dt} = \frac{\theta r}{dt} = \frac{d \theta}{dt} r

v~~~~~~=

\omega r

T = \frac{2\pi r}{v} = \frac{2 \pi}{\omega}

\frac{ds}{dt} = \frac{d \omega r}{dt} = \frac{d \omega}{dt} r

kinematic of circular motion

H&R CH10 rotation

\mathrm{angular~ position} ~~\theta = \frac{s}{r} \frac{[L]}{[L]}

radians

s = \theta r

\frac{ds}{dt} = \frac{\theta r}{dt} = \frac{d \theta}{dt} r

v~~~~~~=

\omega r

T = \frac{2\pi r}{v} = \frac{2 \pi}{\omega}

\frac{dv}{dt} = \frac{d \omega r}{dt} = \frac{d \omega}{dt} r

a~~~~~~=

\alpha r

kinematic of circular motion

H&R CH10 rotation

\mathrm{angular~ position} ~~\theta = \frac{s}{r} \frac{[L]}{[L]}

radians

s = \theta r

\frac{ds}{dt} = \frac{\theta r}{dt} = \frac{d \theta}{dt} r

v~~~~~~=

\omega r

T = \frac{2\pi r}{v} = \frac{2 \pi}{\omega}

\frac{dv}{dt} = \frac{d \omega r}{dt} = \frac{d \omega}{dt} r

a~~~~~~=

\alpha r

a = \frac{v^2}{r} ={\omega^2 r}

kinematic of circular motion

H&R CH10 rotation

kinematic of circular motion

H&R CH10 rotation

kinematic of circular motion

H&R CH10 rotation

kinematic of circular motion

What is the direction of the rotational velocity vector when you ride a bike?

H&R CH10 rotation

K =\frac{1}{2} m_1v_1^2 + \frac{1}{2} m_2v_2^2.... + \frac{1}{2} m_Nv_N^2 = \frac{1}{2}\sum_i m_iv_i^2

energy of rotation

H&R CH10 rotation

K =\frac{1}{2} m_1v_1^2 + \frac{1}{2} m_2v_2^2.... + \frac{1}{2} m_Nv_N^2 = \frac{1}{2}\sum_i m_iv_i^2

K = \frac{1}{2} \sum_i m_ir_i^2 \omega_i^2

energy of rotation

H&R CH10 rotation

K =\frac{1}{2} m_1v_1^2 + \frac{1}{2} m_2v_2^2.... + \frac{1}{2} m_Nv_N^2 = \frac{1}{2}\sum_i m_iv_i^2

K = \frac{1}{2} \sum_i m_ir_i^2 \omega_i^2

moment of inertia

I = \sum_i m_ir_i^2 \\ K = \frac{1}{2}I \omega_i^2\\

energy of rotation

H&R CH10 rotation

K =\frac{1}{2} m_1v_1^2 + \frac{1}{2} m_2v_2^2.... + \frac{1}{2} m_Nv_N^2 = \frac{1}{2}\sum_i m_iv_i^2

K = \frac{1}{2} \sum_i m_ir_i^2 \omega_i^2

I = \sum_i m_i r^2~~\mathrm{or}~~ I = \int r^2 dm \\ K = \frac{1}{2}I \omega_i^2\\

energy of rotation

moment of inertia

system of particles

solid body

H&R CH10 rotation

K = \frac{1}{2} \sum_i m_ir_i^2 \omega_i^2

moment of inertia

I = \int r^2 dm \\ K = \frac{1}{2}I \omega_i^2\\

energy of rotation

H&R CH10 rotation

energy of rotation

K =\frac{1}{2} m_1v_1^2 + \frac{1}{2} m_2v_2^2.... + \frac{1}{2} m_Nv_N^2 = \frac{1}{2}\sum_i m_iv_i^2

I = \int r^2 dm \\ K = \frac{1}{2}I \omega_i^2\\

moment of inertia

the CoM has v=0

K = \frac{1}{2} \sum_i m_ir_i^2 \omega_i^2

H&R CH10 rotation

K =\frac{1}{2} m_1v_1^2 + \frac{1}{2} m_2v_2^2.... + \frac{1}{2} m_Nv_N^2 = \frac{1}{2}\sum_i m_iv_i^2

K = \frac{1}{2} \sum_i m_ir_i^2 \omega_i^2

I = \int r^2 dm \\ K = \frac{1}{2}I \omega_i^2\\

moment of inertia

energy of rotation

H&R CH10 rotation

I = \int r^2 dm \\ K = \frac{1}{2}I \omega_i^2\\

energy of rotation

Parallel Axis Theorem

The moment of inertia of a body is equal to me moment of inertia of the body's center of mass rotation measured along an axis parallel to the axis of rotation, plus the product of the body's mass times the square distance between the axis of rotation and the axis of rotation that passes through the CoM.

I = I_\mathrm{CoM} + Mh^2

H&R CH10 rotation

I = \int r^2 dm \\ K = \frac{1}{2}I \omega_i^2\\

energy of rotation

Parallel Axis Theorem

I = I_\mathrm{CoM} + Mh^2

Let h be the distance between the axis of rotation and the axis parallel to the axis of rotation that goes through the CoM

distance: length of the perpendicular line that joins the two axes

Then the moment of inertia I is equal to

H&R CH9 CoM - momentum

energy of rotation

I = \int r^2 dm = \int [(x - a)^2 + ( y - b)^2] dm

moment of inertia

I = \int r^2 dm \\ K = \frac{1}{2}I \omega_i^2\\

H&R CH9 CoM - momentum

energy of rotation

I = \int r^2 dm = \int [(x - a)^2 + ( y - b)^2] dm

= \int [x^2 + a^2 - 2ax + y^2 + b^2 - 2by] dm =

moment of inertia

I = \int r^2 dm \\ K = \frac{1}{2}I \omega_i^2\\

H&R CH9 CoM - momentum

energy of rotation

I = \int r^2 dm = \int [(x - a)^2 + ( y - b)^2] dm

= \int [x^2 + a^2 - 2ax + y^2 + b^2 - 2by] dm =

= \int (x^2+y^2)~ dm - 2a \int x ~dm - 2b \int y ~dm + (a^2 + b^2) \int dm

moment of inertia

I = \int r^2 dm \\ K = \frac{1}{2}I \omega_i^2\\

H&R CH9 CoM - momentum

energy of rotation

I = \int r^2 dm = \int [(x - a)^2 + ( y - b)^2] dm

= \int [x^2 + a^2 - 2ax + y^2 + b^2 - 2by] dm =

= \int (x^2+y^2)~ dm - 2a \int x ~dm - 2b \int y ~dm + (a^2 + b^2) \int dm

I_{CoM}

moment of inertia

I = \int r^2 dm \\ K = \frac{1}{2}I \omega_i^2\\

H&R CH9 CoM - momentum

energy of rotation

I = \int r^2 dm = \int [(x - a)^2 + ( y - b)^2] dm

= \int [x^2 + a^2 - 2ax + y^2 + b^2 - 2by] dm =

= \int (x^2+y^2)~ dm - 2a \int x ~dm - 2b \int y ~dm + (a^2 + b^2) \int dm

I_{CoM}

moment of inertia

I = \int r^2 dm \\ K = \frac{1}{2}I \omega_i^2\\

CoM coords = 0

H&R CH9 CoM - momentum

energy of rotation

I = \int r^2 dm = \int [(x - a)^2 + ( y - b)^2] dm

= \int [x^2 + a^2 - 2ax + y^2 + b^2 - 2by] dm =

= \int (x^2+y^2)~ dm - 2a \int x ~dm - 2b \int y ~dm + (a^2 + b^2) \int dm

I_{CoM}

M h^2

moment of inertia

I = \int r^2 dm \\ K = \frac{1}{2}I \omega_i^2\\

H&R CH9 CoM - momentum

moment of inertia

I = \int r^2 dm \\ K = \frac{1}{2}I \omega_i^2\\

energy of rotation

I = \int r^2 dm = \int [(x - a)^2 + ( y - b)^2] dm

= \int [x^2 + a^2 - 2ax + y^2 + b^2 - 2by] dm =

= \int (x^2+y^2)~ dm - 2a \int x ~dm - 2b \int y ~dm + (a^2 + b^2) \int dm

I_{CoM}

M h^2

H&R CH9 CoM - momentum

energy of rotation

H&R CH9 CoM - momentum

energy of rotation

CoM_{tr} = (b/2, h/3)

H&R CH9 CoM - momentum

energy of rotation

CoM_{tr} = (b/2, h/3)

H&R CH9 CoM - momentum

energy of rotation

CoM_{tr} = (b/2, h/3)

H&R CH9 CoM - momentum

energy of rotation

r_{CoM}= 1/M\int r dm

r_{CoM}= \rho/M\int_{y_0(x)}^{y_f(x) }\int_{x_0(y)}^{x_f(y)} \sqrt{x^2 + y^2} dx ~dy

CoM_{tr} = (b/2, h/3)

H&R CH9 CoM - momentum

energy of rotation

CoM

energy of rotation

CoM

H&R CH9 CoM - momentum

energy of rotation

CoM

M = 34 g\\ h = 34 ~cm

I = I_{CoM} + 0.34*0.34^2 [kg ~m^2]= 3.9 x 10^{-2}

H&R CH9 CoM - momentum

energy of rotation

I = I_{CoM} + M \left((\frac{3}{4}L \sin{\theta})^2 + (\frac{1}{2}L \cos{\theta})^2\right)

H&R CH9 CoM - momentum

energy of rotation

I = I_{CoM} + M \left((\frac{3}{4}L \sin{\theta})^2 + (\frac{1}{2}L \cos{\theta})^2\right)

Torque is a turning or twisting action on a body about a rotation axis due to a force F

vector quantity
has to do with the force F
has to do with the rotation axis located at some r

torque

Torque is a turning or twisting action on a body about a rotation axis due to a force F

vector quantity
has to do with the force F
has to do with the rotation axis located at some r

\theta

\vec{F}

\vec{r}

\tau = rF \sin{90^o} = rF

torque

\vec{\tau} = \vec{F}\times \vec{r} = r F \sin{\theta} ~[N\cdot m]

Torque is a turning or twisting action on a body about a rotation axis due to a force F

vector quantity
has to do with the force F
has to do with the rotation axis located at some r

\theta

\vec{F}

\vec{r}

\tau = rF \sin{30^o} = \frac{rF}{2}

torque

\vec{\tau} = \vec{F}\times \vec{r} = r F \sin{\theta} ~[N\cdot m]

Torque is a turning or twisting action on a body about a rotation axis due to a force F

vector quantity
has to do with the force F
has to do with the rotation axis located at some r

\vec{F}

\vec{r}

\tau = rF \sin{0^o} = 0

torque

\vec{\tau} = \vec{F}\times \vec{r} = r F \sin{\theta} ~[N\cdot m]

Torque is a turning or twisting action on a body about a rotation axis due to a force F

vector quantity
has to do with the force F
has to do with the rotation axis located at some r

\theta

\vec{F}

\vec{r}

line of action

moment arm

torque

Torque is a turning or twisting action on a body about a rotation axis due to a force F

vector quantity
has to do with the force F
has to do with the rotation axis located at some r

Case Study: Effect of Handrim Diameter on Performance in a Paralympic Wheelchair Athlete -Gabriel Brizuela Costa, et al. 2009

When greater torque is needed, however, [...] a larger diameter could be more effective. Guo et al., (2006) studied the effect of the diameter of the propulsion handrim (0.54 m, 0.43 m, and 0.32 m) on the propulsive moment generated [...] kinetic, potential, and total mechanical energy delivered [...] with the large handrims were significantly greater than with the small ones. Nevertheless the authors conclude that propelling the with the larger handrim size will have a greater metabolic cost.

\vec{\tau} = \vec{F}\times \vec{r} = r F \sin{\theta} ~[N\cdot m]

torque and Force

Torque is a turning or twisting action on a body about a rotation axis due to a force F

vector quantity
has to do with the force F
has to do with the rotation axis located at some r

\vec{\tau} = \vec{F}\times \vec{r} = r F \sin{\theta} ~[N\cdot m]

F_t = ma_t\\ \tau = F_t~r = ma_t r\\

particle of mass m at radius r

torque and Force

Torque is a turning or twisting action on a body about a rotation axis due to a force F

vector quantity
has to do with the force F
has to do with the rotation axis located at some r

F_t = ma_t\\ \tau = F_t~r = ma_t r\\

a_t = \alpha r\\ \tau = F_t~r = ma_t r = \\m (\alpha r) r = m r^2 \alpha

particle of mass m at radius r

\vec{\tau} = \vec{F}\times \vec{r} = r F \sin{\theta} ~[N\cdot m]

torque and Force

Torque is a turning or twisting action on a body about a rotation axis due to a force F

vector quantity
has to do with the force F
has to do with the rotation axis located at some r

F_t = ma_t\\ \tau = F_t~r = ma_t r\\

\alpha = a_t r\\ \tau = F_t~r = ma_t r = \\m (\alpha r) r = m r^2 \alpha

I = mr^2\\ \vec{\tau} = I \vec{\alpha}

particle of mass m at radius r

\vec{\tau} = \vec{F}\times \vec{r} = r F \sin{\theta} ~[N\cdot m]

work and rotational K

Remember:

\Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = W

W = \int_{x_i}^{x_f} F dx

P = \frac{dW}{dt} = F\frac{dx}{dt} = Fv

K = \frac{1}{2}mv^2

K = \frac{1}{2}I\omega^2

For rotation

work and rotational K

Remember:

\Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = W

W = \int_{x_i}^{x_f} F dx

P = \frac{dW}{dt} = F\frac{dx}{dt} = Fv

K = \frac{1}{2}mv^2

K = \frac{1}{2}I\omega^2

\Delta K = \frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_i^2 = W

For rotation

work and rotational K

Remember:

\Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = W

W = \int_{x_i}^{x_f} F dx

P = \frac{dW}{dt} = F\frac{dx}{dt} = Fv

W = \int_{\theta_i}^{\theta_f} \tau d\theta

K = \frac{1}{2}mv^2

K = \frac{1}{2}I\omega^2

\Delta K = \frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_i^2 = W

For rotation

work and rotational K

Remember:

\Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = W

W = \int_{x_i}^{x_f} F dx

P = \frac{dW}{dt} = F\frac{dx}{dt} = Fv

W = \int_{\theta_i}^{\theta_f} \tau d\theta

P = \frac{dW}{dt} = \tau\frac{d\theta}{dt} = \tau\omega

K = \frac{1}{2}mv^2

K = \frac{1}{2}I\omega^2

\Delta K = \frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_i^2 = W

For rotation