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A hollow spherical shell with mass 2.00 $\mathrm{kg}$ rolls without slipping down a $38.0^{\circ}$ slope. (a) Find the acceleration of the shell and the friction force on it. Is the friction kinetic or static friction? Why? (b) How would your answers to part (a) change if the mass were doubled to 4.00 $\mathrm{kg} ?$

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a) 4.8 $\mathrm{N}$ The frictional force is static; it is needed to prevent the body from slipping and to give the ball its angular acceleration.b) The acceleration does not change since it is independent of mass.The friction force is doubled since it depends directly on mass.

Physics 101 Mechanics

Chapter 10

Dynamics of Rotational Motion

Newton's Laws of Motion

Rotation of Rigid Bodies

Equilibrium and Elasticity

Cornell University

University of Michigan - Ann Arbor

Hope College

Lectures

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So let's draw the free by the diagram of the shell. See of this incline with an angle theta. We have this Harlow spherical show. Ah, here we have a force of friction. Going down would be the ah linear acceleration of the center of mass from the center. Going straight down is thief forced to gravity mg and then perpendicular to the surface of Thean Klein is the normal force. And we know that the angle exploration here is going to be in this case counterclockwise. So we know that this is again a hollow spherical show, which means that the moment of inertia is going to be equal to 2/3 m r. Squared. We know that the mass of the shell is two kilograms and that fada equals 38 degrees. So we can say that sigma path in the ex direction will be equal to mass times acceleration of the center of mass. And this will equal mg sign of fada minus the force of friction. Um, we convince say that the sum of torque is going to be equal to the moment of inertia, times the annual exhilaration. So this means that the force of friction Time's rate times the radius this would be equal to torque would be equal. Teo. The Moment of Inertia which we found to be 2 2/3 m r squared times the annual exploration. So to M R Squared Alfa divided by three Solving for Alfa we have Alfa equals the moment of the rather the acceleration of the center of mass divided by our and we find that the force of friction is equaling two times the mass times the linear acceleration of the center of mass divided by three. We can then solve for this and say mg sign of theta minus two times the mass times thie linear acceleration of the center of mass divided by three is going to be equal to the mass times the linear acceleration of the center of mass eso. When we're trying to find the acceleration of the center of mass, we can simply say that this is going to be equal to three g sign of Fada divided by five so we can solve and say three times 9.8 times sign of 38 degrees divided by five and this is equaling 3.62 meters per second. So this is your answer for party. This would be the linear acceleration of the center of mass and then for the force of friction is going to be equal to two times the mass times the acceleration of the center of mass divided by three. So this will be equal to two times two times 3.62 divided by again three. And this is giving us 4.83 Nunes. Now the question is asking us if this Connecticut or Static in this case friction is going to be static, static friction. Because there there is no slipping at the point of contact between the Harlow Spherical show and the incline and then for port be there asking us what happens if the spherical shell was increased to four kilograms. Well, we know that based on the formula for the scent for the linear acceleration of the center of mass, this has no bearing on the mass of the object. Rather, the mass of the object has no bearing on the linear acceleration of the center of mass. So we can say at this acceleration of the center of Mass is independent of mass so we can say that the center of mass of a four kilogram shell will be equal to the acceleration of the center of mass of a two kilogram shell. And again, this is three 0.62 meters per second squared. This will be answered for part B. That is the end of the solution. Thank you for watching.

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