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A solid disk of radius 8.50 $\mathrm{cm}$ and mass $1.25 \mathrm{kg},$ which is rolling at a speed of 2.50 $\mathrm{m} / \mathrm{s}$ , begins rolling without slipping up a $10.0^{\circ}$ slope. How long will it take for the disk to come to a stop?

2.20 $\mathrm{s}$

Physics 101 Mechanics

Chapter 10

Dynamics of Rotational Motion

Newton's Laws of Motion

Rotation of Rigid Bodies

Equilibrium and Elasticity

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Simon Fraser University

University of Sheffield

Lectures

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.

02:21

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.

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A solid disk is rolling wi…

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Well, problem 17 presents us with a good little challenge. Here we have a solid disc with a given radius of 8.5 centimeters. When in a way, the meters and it is rolling. Uh, a slope has a radius of 10 and we're rolling it up. It has an initial speed of the hill here of 2.5 meters per second. And we are asked to find, um What is how long? What's the time? It's gonna take its initial speed to stop. The question is, what time is that going to take? We should be able to see right here. This leads us to a little bit of a kid and kin o Matics formula. If we know our acceleration, we could find our time because we know our initial and final velocities. So the challenge here is to find our acceleration. And so it's kind of look at a free body here, look at what forces we have acting on it. You have a downward force right here. Due to the friction, that is what is hoping to slow that down. Right here at the contact. We have an upward normal horse, and from the center of mass of that solid sphere. We have a force of gravity were told that you I went to 1.25 kilograms. So it's way is well, what's to five minutes? No, we are going to look at it from the perspective that two things Arnett forces in the ex direction caused the acceleration. So if we look at it from there, we know that there some of the forces in the X Emmy there are two forces acting in the ex direction, both actually acting in the negative X direction down the slope. 1st 1 being friction in the 2nd 1 being the horizontal component of the weight of the ball. That would be sign of the 10 degree angle, and that's equal to is to five a. So that's from our Newton's statement. No, nothing going on is it's rolling. So we have some torque going on, and we know our network is moment of inertia times angular exhilaration. Now, remember the relationship between these two guys. The translational is our angular, so angular would be insulation over. Help us out here. Go ahead. Remember, this is a bigger So I left. First, look at what is providing tour? What is their moment of inertia? So it was good a moment of inertia. To begin with, we have a solid solids, beers, moment of inertia is 2/5 Amaar Squared tour being provided by friction. And that's it because the other two forces acting go through the rotation points, so there's no lever arms. So this would be friction times Lever arm, in this case is the radius. And this is 2/5 more squared problems. Hey, conveniently, that helps us out and gets rid of us. Some of our radius is one. And then there's one on each side. So, really, we have solved that friction is two hits. And hey, if we look to put this in right here for friction, we then just consult easily for a So that would be negative. Two hits. Mm, I minus Well, 1 to 5 side John equals 1.25 Hey, chickens of in our mask is we do know our masses 1.25 foraging a little bit of algebra, getting our raids together and solving for a yes, a negative one point two years per second squared, which is what we're looking for? Over here you put zero in 0.5 and negative one. Want to cheat? You go solving qwerty and we get to what, one seconds? Thank you for solving with me today.

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