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A solid, uniform ball rolls without slipping up a hill, as shown in Figure $9.36 .$ At the top of the hill, it is moving horizontally; then it goes over the vertical cliff. (a) How far from the foot of the cliff does the ball land, and how fast is it moving just before it lands? (b) Notice that when the ball lands, ithas a larger translational speed than it had at the bottom ofthe hill. Does this mean that the ball somehow gained energyby going up the hill? Explain!

a) $2.39 \mathrm{s}$36.5 $\mathrm{m}$(b) The total kinetic energy of the ball, just before it lands, is equal to the kinetic energy of the ball at the bottom of the hill. since the rotational kinetic energy of the ball decreases just before it lands, so the translational kinetic energy is greater.

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

Simon Fraser University

University of Winnipeg

McMaster University

Lectures

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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.

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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.

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the first goal. This problem is Steve's conservation of energy to determine what the velocity of the ball is at the top before it launches off the cliff. So we're going to let why go zero be the bottom. So that's the bottom of the cliff. And so why I, which is before it rolls up, is equal to zero, which implies that the initial potential is equal to zero as well. This means that the final position lie f is equal to 20 meters. So if you want zero, this term goes away here and in general, the kinetic energy is equal to 1/2 I omega squared, plus 1/2 film me squared. In this case, we're looking at a solid ball. So I z equal to two fists. Omar squared. Also, we're going to need the fact that the when air velocity is related rotational velocity by our omega since it's rolling without slipping. And so playing these two things in here give 1/2 times two fous and more squared. Omega is equal Toby over our side of the over our square. And then I added to 1/2 of the square to his cancel here are squares. Cancel here side 1/5 film V squared plus 1/2 y b squared. Forget it, Common denominator. For that, you'll find that this is equal to seven tents and V squared. And so the conservation of energy here ultimately tells us that seven tense L'm V I squared is equal to mg y off. Remember, wife is known here, plus seven tense mass, the F squared. And so we can solve this for V f, which is what we want to figure out. And when we do that, we get the P F is equal to square V. I squared Maya's 10 Sevens. Gee, why f and everything here on the right inside is known so we can plug it in and we find the VFW's 15.3 meters per second. Now that's the first step to solving the problem. Now we're going to use some kind of manic equations to solve the rest. We're gonna let plus why be downward and so be not. Why is equal to zero acceleration wise equal to 9.8 positive? Because of this convention I chose here and the distance traveled is equal to 28 meters and So why minus wine on is equal to the nottie, plus 1/2 a T square. I'm going to use this comeback equation to find t. Since B not really to be not why here is equal to zero that term drops out. It's easier to solve for tea now. Teens of being equal to the square of two times in my mind is Why not over, Sergeant? My direction and plugging these values in here give T value of 2.39 seconds. So it takes 2.39 seconds for the mall to hit the ground after launch. Soft Clift. Now we're going to use a different dynamic equation. This one's ex My sex not is equal to be not tea. Really. This comeback equation is just the definition of velocity, but we're going to use it to figure out what X my sex eyes, how far the ball travels on the extraction. We know V not X. That's what we spend a while solving before it's equal of 15.3 meters per second and then t we just all for article 2.39 seconds. So this is equal to 36.6 meters. Now we're going to use the chemical equation V. Why is equal to be Nana? Why plus acceleration times tolling, which is really the definition of acceleration. We're gonna sell this for the bossy in the white direction at the bottom. So this is equal to zero and then we just solved for the time and we know the acceleration. And so this is equal to 9.8 times 2.39 which is equal to 23.4 meters per second. We also need V X, or call that there's no acceleration in this direction. So it's just equal to be not X which was solved for the very beginning. So it's 15.3 meters per second. Now that we have the components of the velocity and the accent, why we can figure out the resulting ultimate velocity Bye Pythagorean theorem. So he is just the square root of the expert. Those of us born looking in these two values here gives us a value of the of 28 meters per second and then see insert to party. Part B is not quantitative. It's purely qualitative, is asking if we just create energy have nothing. And the answer is, of course, no, because the rotation at the top of the hill is less than the rotation, the bottom of the hill. So because the translational kinetic energy is larger after the ball has dropped off the cliff does not mean energy came from nothing. It just means that we have transformed rotational connect energy in the translational Kanak unity, and that's the reasoning behind Part Me, which is the final answer.

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