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A solid uniform spherical boulder rolls down the hill as shown in Figure $9.35,$ starting fromrest when it is 50.0 $\mathrm{m}$ above the bottom. The upper half of thehill is free of ice, so the boulder rolls without slipping. But the lower half is covered with perfectly smooth ice. How fast is the boulder moving when it reaches the bottom of the hill?

29 $\mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 9

Rotational Motion

Physics Basics

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

Cornell University

Simon Fraser University

University of Sheffield

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in this problem. We have a moment of inertia of two fists are squared and we know for the first half of the trip there's gonna be no slipping, which implies that the linear velocity is he going all our times angular velocity. Now, for the section where the motor is on the ice, the rotational connect energy will not change one bit. So we'll be the same for as it is after for this problem, I'm going to let a positive wide the upward I'm gonna let Y is equal to zero at the bottom of the hill. So why is he God was here at the bottom? I'm also gonna let V one be the velocity, the velocity at the end of the rough part at the end of rough part. And so the first gold has promised fear what V one is. Do this. We're going to use conservation of energy which says that the initial kinetic plus the initial potential is equal to final kinetic plus the final potential in this case, the initial Connecticut zero. Since it starts at rest and so continuing along with that expression, we get mg y I, which is the initial potential. This is equal to 1/2 and you one square plus 1/2 I Omega squared. So says the final translational kinetic. This is the final rotational kinetic, and we need to add this with final potential, which is LG. Why now? Let's look at this term here for a second. 1/2 Iomega squared is equal to 1/2 times two fists, you know, more squared, since that's what the moment ownership is. And since omega is equal to feed over our, we give you our spirit like that since it's rolling without slipping. Now the R SK words cancel these twos cancel. And so this just yields a 1/5 on B squared, which combines with this 1/2 don't be square, which gives a seven tense it'll be squared. So I'm going to bring this term here over to the left. And I did him G Why? I'm honest wife and this is equal to seven tense. Tell me one square like that. This expression here can be solved for being one, since we know everything else and when we do that, we get everyone is equal to 18.7 meters per second. This isn't the answer, but it's the first big step toward getting the answer. And so now, for the smooth part, since we've only analyzed the rough part. So the smooth part, the initial velocity is well, we just calculated it's 18.7. The initial height is 25 the final height zero. And so we're going to do the same thing. Conservation of energy. In this case, the final potential zero since ends up at zero right, and the initial Connecticut is 1/2 film V. One squared the initial potential. Actually, there's no more term here. There's the rotational part. There's 1/2 Iomega Square, and the potential here gives mg by and this is equal to K F, which is 1/2 VF square, plus the same rotation apart. Remember at the beginning, I said, the rotational kinetic energy is a constant throughout the smooth part here. So we still have this term. And the nice thing is that this term we'll cancel this term since they're the same in their own officer size of the equal sign. This will let us all for bf in terms of the one in why I, which we both know because we know this. This gives V f is equal to the square. You ve one squared plus two times gravity times. Why I flooding these values in here? Remember, V I is the one It's uploading. Those in there gives us 29 years per second, which is the final answer.

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