Download the App!

Get 24/7 study help with the Numerade app for iOS and Android! Enter your email for an invite.

(II) Using calculus, derive the angular kinematic equations 9 $\mathrm{a}$ and 9 $\mathrm{b}$ for constant angular acceleration. Start with $\alpha=d \omega / d t .$$$\begin{aligned} \omega &=\omega_{0}+\alpha t \quad v=v_{0}+a t \quad[\text { constant } \alpha, a](9 \mathrm{a}) \\ \theta &=\omega_{0} t+\frac{1}{2} \alpha t^{2} \quad x=v_{0} t+\frac{1}{2} a t^{2}[\text { constant } \alpha, a](9 \mathrm{b}) \end{aligned}$$

Get the answer to your homework problem.

Try Numerade free for 7 days

Like

Report

$\omega=\omega_{0}+\alpha t$, $\theta=\omega_{0} t+\frac{1}{2} \alpha t^{2}$

Physics 101 Mechanics

Chapter 10

Rotational Motion

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

Cornell University

University of Washington

McMaster University

Lectures

02:21

In physics, rotational dynamics is the study of the kinematics and kinetics of rotational motion, the motion of rigid bodies, and the about axes of the body. It can be divided into the study of torque and the study of angular velocity.

02:34

In physics, a rigid body is an object that is not deformed by the stress of external forces. The term "rigid body" is used in the context of classical mechanics, where it refers to a body that has no degrees of freedom and is completely described by its position and the forces applied to it. A rigid body is a special case of a solid body, and is one type of spatial body. The term "rigid body" is also used in the context of continuum mechanics, where it refers to a solid body that is deformed by external forces, but does not change in volume. In continuum mechanics, a rigid body is a continuous body that has no internal degrees of freedom. The term "rigid body" is also used in the context of quantum mechanics, where it refers to a body that cannot be squeezed into a smaller volume without changing its shape.

01:47

Using calculus, derive th…

00:20

Solve for the indicated va…

04:43

If angular displacement (…

02:41

The angular displacement …

00:59

The angular velocity of a …

03:40

The angular acceleration $…

03:08

Determine the angular acce…

03:15

(a) Derive Eq. (9.12) by c…

01:46

04:30

The angular acceleration o…

03:32

(II) The angular accelerat…

06:08

The T-shaped body rotates …

So here we're going to say that the angular acceleration by definition, would be the derivative of the angular velocity with respect the time. Ah, and we know that the huh? The initial angular. Rather, we're simply going to say that we're gonna assume that the acceleration is constant. And then we can say that the angular speed at time equals zero is Omega initial. Therefore, we're just starting at T equals zero seconds. And this would just set up thes boundaries for the integral that we need to take. So if we start off by this, we can say that D w we're sorry. The Omega my My apologies would be equal to Alfa DT. At this point, we can integrate, and we can say that the integral from Omega initial to Omega Final of the Omega would be equal to zero to t assuming that tea is starting at zero seconds of Alfa DT. At this point, we can solve. And so this will become Omega final minus Omega initial. This will equal Al Fatih and solving for a mega final. This would simply vehicle to make initial plus offer tea. So that's how we derive this equation. now. Furthermore, we know that we can say that Omega Final would be equal to Omega Initial plus Al Fatih. So just rewrite this. However, this is gonna equal the change in angular displacement. With respect to time according to the definition of angular velocity we can then odds break algebraic Lee, manipulate this and say, D theta well equal Omega initial plus off the tee times again D t. And now we have to integrate once again integrate from zero to theta of de fada. Uh, this would be equal to the integral from zero to t for omega initial plus Alfa t times. And we find that Fada the change and angular displacement would be equal to Omega Initial times t plus 1/2 times outfit T Square. So this would be how we were to derive the equation for the change in angular displacement. That is the end of the solution. Thank you for watching

View More Answers From This Book

Find Another Textbook

Numerade Educator