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Doc Al

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His mistake is not realizing that the point of contact is accelerating and thus different rules apply.Then he drew a diagram and a gravity vector going straight down the center of the sphere and said gravity provides torque if we considerthe point of contact as the pivot.

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Can you please elaborate as to what mathematical statement can be used as an evidence that the ball will definitely slide and not spin about it COM. His thinking doesn't really matter, as long as I understand what is happening.You are correct and your teacher is wrong. If there is no friction, there will be no torque about the center of mass to produce a rotational acceleration.

His mistake is not realizing that the point of contact is accelerating and thus different rules apply.

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Doc Al

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If he chooses to use the point of contact as the reference point, that's fine, but a more advanced analysis is required: The bottom line is that a proper analysis using that

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Nugatory

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That was my first thought, but then I found I myself thinking about the normal force exerted by the surface of the inclined plane.... Does it pass through the center of mass of the sliding object? If I were to try standing on a frictionless incline looking downhill, my feet would go out from under me and the back of my head would hit the surface.You are correct and your teacher is wrong. If there is no friction, there will be no torque about the center of mass to produce a rotational acceleration.

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Doc Al

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For spheres and rings, yes. (That's what we're talking about here. I think!)That was my first thought, but then I found I myself thinking about the normal force exerted by the surface of the inclined plane.... Does it pass through the center of mass of the sliding object?

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Thank you for your time, my query feels answered. What textbooks would you recommend for one level above the introductory level, as I would like to step into physics at home.alwaysuse the COM as your reference point. Since there is no torque about the COM, there will be no rotational acceleration. Simple as that.

If he chooses to use the point of contact as the reference point, that's fine, but a more advanced analysis is required: The bottom line is that a proper analysis using thatacceleratingreference point will also show that there will be no rotational acceleration about the COM. (This is covered in every Classical Mechanics textbook. It's not really that advanced, just one step above Intro physics.)

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Ah - yes, of course. I posted, went off to do start doing the dishes, and then realized that we're talking about circular spheres and rings so a vector perpendicular to the surface will pass through the center. I suspect that OP's teacher made the same mistake, failing to recognize the the importance of the symmetry in this case.For spheres and rings, yes. (That's what we're talking about here. I think!)

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Mister T

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You can easily simulate the situation you proposed in your previous post by standing on roller skates on a ramp or hill. If you stay on your feet a line perpendicular to the ramp will pass through your feet and your center of mass, something that most roller skaters can manage, and then you'll move down the ramp without rotating. On the other hand, if you fell over instead, you would still slide down the frictionless ramp without rotating.Ah - yes, of course. I posted, went off to do start doing the dishes, and then realized that we're talking about circular spheres and rings so a vector perpendicular to the surface will pass through the center.

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A.T.

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He is wrong as Doc Al wrote. Here is another way to think about it: A uniform gravitational field alone never makes stuff spin, only the non-uniformly applied contact forces do, if they create a torque around the COM.Then he drew a diagram and a gravity vector going straight down the center of the sphere and said gravity provides torque if we consider the point of contact as the pivot.

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CWatters

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Draw a free body diagram of the ball.Then he drew a diagram and a gravity vector going straight down the center of the sphere and said gravity provides torque if we consider the point of contact as the pivot.

There is no friction so the reaction force between the ball and slope must be normal to the slope

1) It's useful to remember this for other situations such as ladders leaning against frictionless walls.

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Just another take on this. Simply decompose the gravitational force into normal and tangential components. The normal component cancels the normal force from the slope, leaving simple tangential acceleration.

When there is friction, you have an additional force up the slope. Note that this not only provides the torque for rotation, but also reduces the tangential acceleration. In your teacher's analysis, therefore, the energy would not add up. The rotational energy would come with no reduction to linear kinetic energy.

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But, shouldn't that be the case anyway? That friction is necessary to BEGIN rolling but once you start rolling, the instantaneous velocity of contact point w.r.t. ground is zero. Hence friction should not cost any energy as it does ZERO work. Am I right?

Just another take on this. Simply decompose the gravitational force into normal and tangential components. The normal component cancels the normal force from the slope, leaving simple tangential acceleration.

When there is friction, you have an additional force up the slope. Note that this not only provides the torque for rotation, but also reduces the tangential acceleration. In your teacher's analysis, therefore, the energy would not add up. The rotational energy would come with no reduction to linear kinetic energy.

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In the case of friction it does zero work. The total kinetic energy of the sphere, rotational plus linear, equals the loss of gravitational potential energy. Friction has the effect of causing some of the energy to be rotational - at the expense of linear kinetic energy.But, shouldn't that be the case anyway? That friction is necessary to BEGIN rolling but once you start rolling, the instantaneous velocity of contact point w.r.t. ground is zero. Hence friction should not cost any energy as it does ZERO work. Am I right?

This scenario is worth studying.

But, if a sphere starts to roll without friction, then there is nothing to reduce the linear acceleration. So, all the potential energy must go to linear kinetic energy and the rotational energy must come for free.

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A.T.

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Static friction dissipates no energy. Kinetic friction and rolling resistance do.But, shouldn't that be the case anyway? That friction is necessary to BEGIN rolling but once you start rolling, the instantaneous velocity of contact point w.r.t. ground is zero. Hence friction should not cost any energy as it does ZERO work. Am I right?

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jbriggs444

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Yes, if one uses the point of contact as the [instantaneous] axis of rotation then the force of gravity amounts to an unbalanced torque. However, one must realize that not all angular momentum consists of rotation. In this case, the resulting angular momentum increase is accounted for by the

When the center of mass of an object is moving on a line that does not pass through the chosen axis then that motion has non-zero angular momentum given by ##\vec{L}=\vec{r}\times\vec{p}## where L is the angular momentum, r is the offset of the center of mass from the axis of rotation and p is the linear momentum of the object. If the object accelerates, its linear momentum is increasing along with the associated angular momentum.

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A.T.

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@adesh123 I think the above is the best general explanation. You could just as well drop the ball vertically and calculate a non-zero gravity torque around some random point off the vertical. That torque will not make the ball spin during the fall, even tough it will change the angular momentum around that point.

Yes, if one uses the point of contact as the [instantaneous] axis of rotation then the force of gravity amounts to an unbalanced torque. However, one must realize that not all angular momentum consists of rotation. In this case, the resulting angular momentum increase is accounted for by thelinearacceleration of the sphere or ring.

When the center of mass of an object is moving on a line that does not pass through the chosen axis then that motion has non-zero angular momentum given by ##\vec{L}=\vec{r}\times\vec{p}## where L is the angular momentum, r is the offset of the center of mass from the axis of rotation and p is the linear momentum of the object. If the object accelerates, its linear momentum is increasing along with the associated angular momentum.

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CWatters

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jbriggs444

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Except, of course, for the fact that there

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Suppose I don't know if the ball will slide or spin, what mathematical piece of information would lead me to a conclusion? I'm more of an analytical person than imaginative. It is true that the angular momentum can be provided by linear motion, but what evidence states that it will slide. (P.S. I'm not being in denial or over-skeptical, just want to learn how to go about things in physics)Except, of course, for the fact that thereisa torque on the ball about the point of contact.

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jbriggs444

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There is no torque about its center of mass (on which point I agree with @CWatters). So it will not spin.Suppose I don't know if the ball will slide or spin, what mathematical piece of information would lead me to a conclusion?

More generally, it will slide if its linear acceleration exceeds its rotational acceleration multiplied by its radius. Since its rotational acceleration is zero and its linear acceleration is non-zero, this condition is trivially met. You would typically determine the two accelerations by drawing free body diagrams, determining relevant forces and solving the resulting equations.

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The key piece of information is the maximum friction force.Suppose I don't know if the ball will slide or spin, what mathematical piece of information would lead me to a conclusion? I'm more of an analytical person than imaginative. It is true that the angular momentum can be provided by linear motion, but what evidence states that it will slide. (P.S. I'm not being in denial or over-skeptical, just want to learn how to go about things in physics)

As I mentioned in an earlier post, you should study a sphere and/or a cylinder moving down a frictional slope.

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CWatters

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A free body diagram of the ball has no "point of contact", just the ball and the forces acting on it. No?Except, of course, for the fact that thereisa torque on the ball about the point of contact.

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jbriggs444

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A free body diagram used for computing torque most certainly has the points where forces act and where the axis of rotation is assigned.A free body diagram of the ball has no "point of contact", just the ball and the forces acting on it. No?

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