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(II) A thin, hollow 0.545 -kg section of pipe of radius 10.0 $\mathrm{cm}$ starts rolling (from rest) down a $17.5^{\circ}$ incline 5.60 $\mathrm{m}$ long. (a) If the pipe rolls without slipping, what will be its speed at the base of the incline? (b) What will be its total kinetic energy at the base of the incline? (c) What minimum value must the coefficient of static friction have if the pipe is not to slip?

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(a) 4.06 $\mathrm{m} / \mathrm{s}$(b) $8.99 \mathrm{J}$(c) 0.158

Physics 101 Mechanics

Chapter 10

Rotational Motion

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

Cornell University

Simon Fraser University

Hope College

Lectures

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.

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.

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apply for part of the the the conservation of energy or mechanical energy. So energy initial equals energy final. And we get that the potential energy initial equals the final kinetic energy of the center of mass plus the rotational kinetic energy we can say here mg d sign of fada equals 1/2 times the mass times velocity at the bottom squared plus 1/2 times the moment of inertia I times themoment of times, the angular velocity at the bottom squared. We can then say that, um m g d sign of theta This is gonna equal 1/2 and fi bottom squared. This would be plus 1/2 times moment of inertia M r squared, multiplied by the velocity at the bottom divided by are, uh, quantity squared. Essentially, um, this would be equal to the mass times the velocity at the bottom squared and so for parted The velocity at the bottom will be equal to the square root of G d. Sign of data and we can solve. So this would be the square root of 9.80 meters per second squared multiplied by 5.60 meters, multiplied by sine of 17.5 degrees and so this is equaling 4.6 meters per second. This would be our answer for part A for part B the total kinetic energy at the base of the incline as the same as the initial potential energy. So this would be, um g d sign of Fada we can solve. This would be a point 545 kilograms multiplied by 9.80 meters per second squared multiplied by five 0.60 meters times sign of again 17.5 degrees and we find that the final kinetic energy at the bottom, the total final kinetic energy would be equaling 2.8 point 99 jewels. This would be your answer for part B for part C. We then know that the frictional force supplies the torch for the object to roll without slipping, and the frictional force has a maximum value. So we can say that the object rolls without slipping. We know that the angular acceleration of the equal to the, um, linear acceleration divided by our and we're gonna use Newton's second law for the directions parallel and perpendicular to the plane. And then we're gonna a self. We're going to rather apply the sum of the torques interest to solve for the coefficient of kinetic friction. And so we can start the coefficient of static friction. And so we can say that here, the sum of two works would be equaling force of friction. Times are this would be equaling the moment of inertia, times the angular acceleration. This would be, um, r squared times a divided by our. So this is equaling the mass times. The linear acceleration times are we then know that the force of friction is going to be equaling the mass times acceleration and the sum of forces perpendicular would be equal to the forced normal minus m g co sign if ada on. So we had this really cool zero. So we know that the forced normal is an equaling MG co sign of data. We know that the sum of forces parallel with the equaling mg sign of Fada minus the force, the friction and this is equaling the mass times acceleration. And so the force of friction would be equaling 1/2 times m g sign of Fada. And now we know that four part see the force of friction must be less than or equal to the force of friction. Static Max. And essentially, we would have won half m. G. Sign of Fada is going to be less than or equal to the coefficient of static friction times the force normal where this is equaling the coefficient of static friction. Times MG co sign of Fada on. So we know that the coefficient of static friction must be greater than or equal to 1/2 times tangent of Fada. And so the coefficient of static friction minimum would be equal in 1/2 times tangent of 17.5 degrees. This is equaling 0.158 So the minimum coefficient of static friction would be 0.158 That is the end of the solution. Thank you for

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