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A certain type of propeller blade can be modeled as a thin uniform bar 2.50 m long and of mass 24.0 $\mathrm{kg}$ that is free to rotate about a frictionless axle perpendicular to the bar at its midpoint. If a technician strikes this blade with a mallet 1.15 $\mathrm{m}$ from the center with a 35.0 $\mathrm{N}$ force perpendicular to the blade, find the maximum angular acceleration the blade could achieve.

2.4 $\mathrm{rad} / \mathrm{s}^{2}$

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

Hope College

University of Winnipeg

McMaster University

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hello, Problem number seven models a propeller blade as a thin uniformed bar. That's that is 2.5 meters long. It has a mass of 24 kilograms. It's mass since it's a uniform bar. Its center mass would be directly in the center at one point. Top 25 kilogram are meters from the pivot point right here, which would make its weight be 24 times 9.8. This propeller blade is struck with a mallet at a location 1.15 from the center. In the force that we strike this mallet or this propeller blade with is 35 mins, were asked to find the maximum angular acceleration that the blade would receive. So we now by definition of Twerk, that tor causes angular acceleration. So the sum of all the torques acting on this propeller blooded would be equal to the moment of inertia, times the angular acceleration. There are two different forces that would provide torques. The weights would provide a clockwise tour and would be considered negative. Where they force from. The mallet would provide a counterclockwise force, and it would be a torque and it would be considered positive when we find each individual torque, we do talk times, lever arm, so we have to force or two torques. Acting one is 35. It's lever arm would actually be in its from our pivot 350.1 point 25 plus one point warm far, and the force of gravity would provide torque in the opposite direction. Uh, and it's lever on would be 1.25 Those would be equal to the moment of inertia. Times are angular acceleration. Now this is a thin rod moment of inertia for Thin Rod is 1 12 M R. Squared or 1 12 are Masses 24 and our distance from the pivot point would again be 1.25 plus 1.15 That's 2.4 squared times the angular acceleration solving for angular acceleration. We would get 3.22 radiance. Her second square. Thank you for learning with me today.

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