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(II) A thin rod of mass $M$ and length $\ell$ is suspended vertically from a frictionless pivot at its upper end. A mass $m$ of putty traveling horizontally with a speed $v$ strikes the rod at its $\mathrm{CM}$ and sticks there. How high does the bottom of the rod swing?

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$\frac{m^{2} v^{2}}{g(m+M)\left(\frac{4}{3} M+m\right)}$

Physics 101 Mechanics

Chapter 11

Angular Momentum; General Rotation

Moment, Impulse, and Collisions

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

University of Michigan - Ann Arbor

Simon Fraser University

Hope College

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.

04:12

In physics, potential energy is the energy possessed by a body by virtue of its position relative to others, stresses within itself, electric charge, and other factors. The unit for energy in the International System of Units is the joule (J). One joule can be defined as the work required to produce one newton of force, or one newton times one metre. Potential energy is the energy of an object. It is the energy by virtue of an object's position relative to other objects. Potential energy is associated with restoring forces such as a spring or the force of gravity. The action of stretching the spring or lifting the mass is performed by a force which works against the force field of the potential. The potential energy of an object is the energy it possesses due to its position relative to other objects. It is said to be stored in the field. For example, a book lying on a table has a large amount of potential energy (it is said to be at a high potential energy) relative to the ground, which has a much lower potential energy. The book will gain potential energy if it is lifted off the table and held above the ground. The same book has less potential energy when on the ground than it did while on the table. If the book is dropped from a height, it gains kinetic energy, but loses a larger amount of potential energy, as it is now at a lower potential energy than before it was dropped.

04:40

(II) A thin rod of mass $M…

02:48

A thin uniform rod of mass…

02:33

(a) A thin rod of length $…

01:59

A uniform rod of length $L…

03:33

A thin rod of length $h$ a…

02:38

Thin Rod of length $L$ A t…

01:12

A thin rod of length $0.75…

04:42

A uniform rod $P Q$ (of ma…

01:06

A uniform rod of mass $m$ …

A uniform rod of mass $\ma…

02:20

A thin uniform rod $A B$ …

00:58

A thin uniform rod of leng…

02:09

(II) A uniform horizontal …

you know that we're going to apply the conservation of angular momentum before and after the collision so we can say L before collision will be equal to the angular momentum after the collision. And so we can say that, um, times all over two times V. This would be equaling the moment of inertia of the rod, plus the moment of inertia of the party multiplied by the angular velocity. And so we can see that the angular velocity would be equaling to the mass times the length times V divided by two times the moment of inertia for the Rod Cluster moment of your chef to the party. Ah, we're going to then apply the Newtons started Nunes. We're gonna apply the law of conservation of energy before and after a rather after the collision and the top of the swing. So we can say that the energy after the collision would be equal to the energy at the top of the swing. And so we can say that essentially, the kinetic energy after the collision will be equal to the potential energy at the top of the swing. And so we can say 1/2 times the moment of inertia of the Rod, plus the moment of inertia of the party ah, multiplied by Omega Squared, which we had found previously. This would be equal to the sum of the masses multiplied by GH. And so we can say that the heights of the center of mass would be equal in the moment of inertia for the rod, plus the moment of inertia for the party multiplied by omega squared, the angular velocity squared, divided by two times the's. Some of the masses times the acceleration due to gravity and weaken Now substitute. The height of the center of mass would be equaling. Aah! This would be the moment of inertia for the rod, plus the moment of inertia for the party divided by two g times the sum of the masses and then multiplied by the mass. How the divided by two times the moment of inertia the rod, plus the moment of inertia. Philip Party. This will be squared and so this would be equal. Two. I'm squared elsewhere, B squared, divided by eight g times of some of the masses multiplied by the sum of the moments of inertia for the rod and the party. And so we can then say that we can substitute now for the moment of inertia for the rod and the moment of their chauffeur, the putty. And we can say then so the nominee, er eight g times of some of the masses multiplied by M l squared over three plus damn times out over, too. Quantity squared. Let's This is specifically capital them. And this is Laura. He's, um, and then for the numerator We have, um, squared, all squared d squared. And so this is equaling. I'm squared V squared, divided by two g times to some of the masses multiplied by 4/3 times. Um, plus lower case m. And so we find that the height of the bottom would be equaling two times the height of the center of mass. And this is equaling. I'm squared V squared. And then essentially this to cancels out with this, too. And we have g divided by the sum of the masses multiplied by 4/3. And plus, this would be our final answer. That is the end of the solution. Thank you. For what

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