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(III) A thin rod of length $\ell$ stands vertically on a table. The rod begins to fall, but its lower end does not slide. (a) Determine the angular velocity of the rod as a function of the angle $\phi$ it makes with the tabletop. (b) What is the speed of the tip of the rod just before it strikes the table?

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(a) $\omega=\sqrt{\frac{3 g}{l}(1-\sin \phi)}$(b) $\sqrt{3 g l}$

Physics 101 Mechanics

Chapter 10

Rotational Motion

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

University of Washington

University of Winnipeg

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.

02:32

A long, thin rod of mass $…

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A long, thin rod of mass 8…

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'thin uniform rod of …

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a long, thin rod that is s…

01:52

A uniform rod of mass $M=2…

08:10

15:17

A slender rod of mass $m$ …

05:12

A thin rod (length 1.50 m…

13:06

A slender rod of length $l…

07:20

A uniform rod of length $l…

So here we're gonna, uh, draw the free body diagram for the falling rod. Let's You're our axes. This would be the rod here. Uh, at the center of mass would be the force of gravity. The rod would have the length of l here would be forced normal. This would be force of friction. And we can see that this is an angle Phi. And so we can For part A, we can say that the axis of rotation would be coming out of the paper at the point of contact with the floor. Um, we could say that here on Lee, gravity actually creates torque. And then we can simply say that Ah, we're gonna to apply to ignore to evaluate the network. We can say that counterclockwise is positive. So we can say that the sum of the talks is gonna equal the moment of inertia times the angular acceleration this equal negative and g times 1/2 times l co sign of state us. That would be the length. And then we can simply solve and say that the torque would then be equal to 1/3 times. Um l squared times. The change in angular acceleration, which would be the derivative of the angular velocity. With respect to time, we can then say that negative three g over to l co sign Ah fei with equal the derivative of the angular velocity with respect to time And this would be equal to the derivative Lee A angular velocity with respect to the change and angle multiplied by the change in the angle with respect to time. So then we can rewrite this and say that Okay, uh, three native three g over to El Co sign If I would then be equal to the change in the river the derivative of the angular velocity to fi times the angle fire rewrite and say three g over to L Times Co sign of five times Defy would be equal to negative negative the negative one times the angular velocity Times de omega. Now we integrate so we can say three g over to O integrating from pi over too 225 times co sign if I defy, would be equal to negative times the integral from zero to Omega Times, Omega de Omega and this would give us three g over to L Times. Sign of five minus one. This would equal negative. Omega squared over too. And so Omega is simply gonna be equal to the square root of three G over L times one minus sign of five. This would be our final answer for part a. Now, for part B, we want to find, um, the velocity and the rather the linear velocity at zero at fi equaling zero. So we can say that V is gonna equal the angular velocity. The linear velocity equals the angular velocity times l and this would simply be equal to square root of three g o one minus sign of fi. So when we want the linear velocity at fi equaling zero degrees, this is simply gonna be equal to the square root of three g o. This would be our answer for part B. That is the end of the solution. Thank you for watching

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