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(II) Assume that a $1.00-\mathrm{kg}$ ball is thrown solely by the action of the forearm, which rotates about the elbow joint under the action of the triceps muscle, Fig. $52 .$ The ball is accelerated uniformly from rest to 8.5 $\mathrm{m} / \mathrm{s}$ in $0.35 \mathrm{s},$ at which point it is released. Calculate $(a)$ the angular acceleration of the arm, and $(b)$ the force required of the triceps muscle. Assume that the forearm has a mass of 3.7 $\mathrm{kg}$ and rotates like a uniform rod about an axis at its end.

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(a) 78 $\mathrm{rad} / \mathrm{s}^{2}$(b) 670$N$

Physics 101 Mechanics

Chapter 10

Rotational Motion

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

Cornell University

Hope College

University of Sheffield

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.

06:17

Assume that a $1.00-\mathr…

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(II) Assume that a 1.00-kg…

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(II) Assume that a $1.00-\…

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Assume that a 1.00-kg ball…

04:28

Assume that a 1.20-kg ball…

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Assume that = 30 ~kg ball …

02:32

(II) The forearm in Fig, 5…

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The forearm in Fig. $10-52…

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(II) The forearm in Fig. 8…

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In Figure $8.4 \mathrm{~b}…

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The $0.5-1 b$ ball is guid…

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A baseball pitcher throws …

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BIO Bowling At the start o…

07:51

A thin, uniform, $3.80-\ma…

for party. We need to find the angular acceleration, the definition of the angular acceleration, to be the change in angular velocity divided by the change in time. And so this would be equaling omega time of over tea plugging in for omega. This would be the linear velocity divided by the radius, divided by t. And so this would be equaling 8.5 meters per second, divided by 0.31 meters. This would be divided by t of 0.35 seconds and we find that Alfa is equaling at 78.34 radiance per second squared. This would be our answer for part A for part B. Then we know that the force required can be found from the torque torque equaling f times our sign of data. And we know that here in this situation, the force is perpendicular to the lever arm. So fate as equaling 90 degrees. And we know that the torque is also equaling. Generally the moment of inertia, times the angular acceleration and so equating these two equations, we have our equaling I alfa, and we can then say that the force is gonna be equaling the moment of inertia times the angular acceleration divided by our and we have the mass of the ball times the distance of the ball squared, plus 1/3 times the mass of the lever arm times the length of the lever arm squared, divided by our times Alfa. And so we can say that the force after four partied the force after plugging in would be 1.0 kilograms. The mass of the ball times the distance from the axis of rotation of the ball. 0.31 meters quantity squared, plus 1/3 times, 3.7 kilograms. The mass of the lever arm times 0.31 meters quantity squared the length of the lover arm divided by 0.25 meter the radius and then multiplied by the angular acceleration. 78 point 34 radiance per second squared. And we find that the forces equaling approximately 670 Newton's. This would be our answer for part B. That is the end of the solution. Thank you for

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