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A basketball (which can be closely modeled as a hollowspherical shell) rolls down a mountainside into a valley andthen up the opposite side, starting from rest at a height $H_{0}$ above the bottom. In Figure $9.33,$ the rough part of the terrainprevents slipping while the smooth part has no friction.(a) How high, in terms of $H_{0},$ will it go up the other side?(b) Why doesn't the ball return to height $H_{0} ?$ Has it lost any ofits original potential energy?

a) $\therefore$ The ball goes to the height $\frac{3 H_{o}}{5}$ on the smooth surfaceb) Because it lost some potential energy to overcome the friction of the rough surface.

Physics 101 Mechanics

Chapter 9

Rotational Motion

Physics Basics

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Simon Fraser University

University of Sheffield

Lectures

04:16

In mathematics, a proof is a sequence of statements given to explain how a conclusion is derived from premises known or assumed to be true. The proof attempts to demonstrate that the conclusion is a logical consequence of the premises, and is one of the most important goals of mathematics.

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.

01:09

A basketball (which can be…

07:26

03:49

09:56

A solid uniform ball rolls…

08:13

03:22

Starting from rest, a bask…

13:54

A child rolls a 0.600 -kg …

03:32

04:20

A ball with an initial vel…

In this problem moment, inertia of the basketball is equal to 2/3. No more squared, since it's a hollow, spherical shell. For the rough part, we know V is equal to our omega, which is a relationship the objects abide by when they roll without slipping. We're gonna let plus why be upward, Oregon Lentil it? Why is equal to zero correspond to Valley? No. So the first episode here, what the velocity is in the Valley after the ball is drops a bit? Serena's Conservation of Energy, which says that the initial kinetic was the initial potential, Is he going to the final kinetic? That's the final potential. Now Y F is equal to zero, since Y is equal to zero in the valley, so you f is equal to zero. Likewise, the initial kinetic energy, since it starts from rest, is equal to zero. And so we have you eyes equal to K f, and now a plug in and we get l ght. Since why I is age nine and K F is equal to 1/2 it'll be squared, which is the linear connect energy class. 1/2 Iomega squared, which is the rotational connect energy. This term here is equal to 1/2 times 2/3. You know, Omar squared. I'm just like you. And I know I'm gonna plug in Omega, which is be over our because during this portion, it's rolling without slipping. So we have that relationship between the angle velocity and a linear velocity. This simplifies here to be equals to 1/3 and we scored. And so this 1/3 Envy Square combines with this 1/2 and b squared and a simplifies the whole equation to be MGH not is equal to 56 it'll be squared, which means that B squared is equal to six g h not over five. And we're going to set this to the side for a second. Now, we're going to use conservation of energy where the initial position is in the valley. In the final position is up smooth, part of the hill. And so again we have K II plus you. I is equal to K f plus U f in this case, you wise eagle to zero. Since we're starting in the valley at a height of zero, it's okay. I is 1/2 film V. I squared, which is just this. I'm gonna call it be scored. Plus the rotation part 1/2 I Omega Square. This is equal to the final connect energy, which is equal to 1/2. I'll make squared. There's no translational party more since the final A party will be at rest at the top, but there will still be a rotation associated with it and then a final potential energy. Well, I guess the height age. So that's what that isthe now. These terms cancel says they're on opposite sides of the equal sign and we can solve for H here we get that H is equal to V squared over to G. And we found that beast. Where was it supplying that in? And we get that this is equal to 3/5 h not. And so it achieves less height than you started with and the reason and now we're going to the party. Now the reason this is 3/5 h not is because a lot of the initial potential energy got converted into rotational connect energy. And so this value is less than four, since it has some connect energy associative rotation and that's it

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