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(III) A hammer thrower accelerates the hammer (mass $=7.30 \mathrm{kg}$ ) from rest within four full turns (revolutions) and releases it at a speed of 26.5 $\mathrm{m} / \mathrm{s}$ . Assuming a uniform rate of increase in angular velocity and a horizontal circular path of radius $1.20 \mathrm{m},$ calculate $(a)$ the angular acceleration, $(b)$ the (linear) tangential acceleration, $(c)$ the centripetal acceleration just before release, $(d)$ the net force being exerted on the hammer by the athlete just before release, and $(e)$ the angle of this force with respect to the radius of the circular motion. Ignore gravity.

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(a) 9.7 $\mathrm{rad} / \mathrm{s}^{2}$(b) 11.6 $\mathrm{m} / \mathrm{s}^{2}$(c) 585 $\mathrm{m} / \mathrm{s}^{2}$(d) 4270$N$(e) $1.14^{\circ}$

Physics 101 Mechanics

Chapter 10

Rotational Motion

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

University of Washington

University of Winnipeg

McMaster University

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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have the diagram of the system. And ah, we can say that four part of the angular acceleration can be found from equation 10 9 See. And so we can say that Omega squared would be equal in Omega initial squared plus two alpha Delta Fada and so Alfa. The angular acceleration would be equal into the final angular velocity minus the initial angular velocity divided by two times the change in the angular displacement. And so this would we know the the initial angular velocity zero. And so this would be equaling V over r quantity squared, divided by two times Delta Fada and so Alfa would be equaling 26.5 meters per second, divided by the radius of 1.20 meters quantity squared, divided by two times Delta theta eight pi radiance. And we find that the angular acceleration is equaling 9.70 radiance per second squared. Now, um, we can say for part B to find the tangential acceleration. This would simply be equaling the angular acceleration times the radius are. And so this would be 9.702 We will round at the very end radiance per second squared multiplied by 1.20 meters. This is equaling 11.64 meters per second squared. And so we can say that the centripetal acceleration or we can say For Parsi, the radial acceleration would be equaling V squared over r and so this would be 26.5 meters per second quantity squared, divided by 1.20 meters. And this is equaling 585 0.2 meters per second squared. So this would be your answers for B and C weaken safer Part D. Now the net force is the mass times the nets acceleration so we can say force net would be equaling the mass times the nets acceleration and this would be equaling the mass times the square root of the tangential acceleration square plus the radio acceleration squared. This would be equaling. We could say the net force would be equaling 7.30 kilograms multiplied by the square root of 11.64 meters per second quantity squared plus 585.2 meters per second quantity squared, and we find that the net force is equal in approximately 4270 Newton's. This would be your answer. Four Part D and then four Part E. We're going to find the angle from the two acceleration vectors. So Fada would be equaling the Ark 10 of the tangential acceleration divided by the radio acceleration. And this is equaling Art tan of 11.64 divided by 585.2, and we find that the angle is equaling 1.14 degrees. This would be our answer for Part E. That is the end of the solution. Thank you for

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