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(II) The forearm in Fig, 52 accelerates a 3.6 -kg ball at 7.0 $\mathrm{m} / \mathrm{s}^{2}$ by means of the triceps muscle, as shown. Calculate (a) the torque needed, and (b) the force that must be exerted by the triceps muscle. Ignore the mass of the arm.

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(a) 7.8$m . N$(b) 310$N$

Physics 101 Mechanics

Chapter 10

Rotational Motion

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

Cornell University

University of Michigan - Ann Arbor

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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(II) The forearm in Fig. 8…

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The forearm in Fig. $10-52…

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The forearm in the figure(…

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ConstantsThe forearm i…

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(II) Assume that a $1.00-\…

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(II) Assume that a 1.00-kg…

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Assume that a 1.20-kg ball…

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In Figure $8.4 \mathrm{~b}…

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Assume that a $1.00-\mathr…

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(II) $(a)$ Calculate the f…

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(II) $(a)$ Calculate the m…

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(II) (a) Calculate the mag…

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(I) Approximately what mag…

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A bowler holds a bowling b…

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10:29

so here for part A. We know that the torque gives angular acceleration to the ball on Lee because the arm, which is which with which it is thrown, is considered massless so we can find the angular acceleration of the ball. Um, we confound find the angular acceleration from the ball from the given tangential acceleration. So we can say that torque would be equal to the moment of inertia times the angular acceleration we're gonna We're gonna consider a ball Ah, a distance of are away from the ball, has a mass and is a distance are away from the axis of rotation. So we can say that this would be equal to M r squared times Alfa. This would be equal to m r squared times the 10 gentle acceleration divided by our so this would be equal to m r times the tangential acceleration on Then we can solve so torque ah would be equal to 3.6 kilograms multiplied by 0.31 meters multiplied by the tangential acceleration of 7.0 meters per second squared. This is giving us 7.81 Newton meters now for part B, the triceps muscle must produce the torque required. However, the lever aren't with but the lever arm of on Lee about two points five centimeters. So we can say that the torque here would be equal to half times the perpendicular distance from the axis of rotation. So in order to solve for the force required, we can simply say this would be equal to the torque divided by the perpendicular distance of 2.5 centimeters. So this would be equal to seven point 812 Newton meters, divided by 0.0 to 5 meters, two and 1/2 or two and 1/2 centimeters. We find that the force required is approximately 310 Newtons. We have to round 22 significant figures. That is the end of the solution. Thank you for watching.

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