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Kinesiology 2241A/B Lecture Notes - Lecture 9: Soleus Muscle, Angular Velocity, Angular Acceleration

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marooncicada960
26 Oct 2017
School
Western University
Department
Kinesiology
Course
Kinesiology 2241A/B
Professor
Thomas Richard Jenkyn

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Document Summary

A(cid:272)(cid:272)ordi(cid:374)g to (cid:374)e(cid:449)to(cid:374)"s 2nd law: when a force is applied to an object then it accelerates in a straight line, the force must be applied at the center of the mass. If the object is a body segment that is attached at one end via a joint, then the rotation (and the torque) acts about the joint. When the gastrocnemius and soleus muscles pull on the achilles tendon, the foot. Segment rotates about the ankle (and not the foot center of mass) So torque (aka moment of force) has a magnitude of: torque = force x perpendicular distance. The angular acceleration that results from tis torque is: t = i x alpha. Where i is the moment of inertia (ie, rotational mass) and alpha is the angular velocity. Just like a force, torque is a vector and so it has the same 4 characteristics of any vector: 1. Magnitude is t= f x d: 2.

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Related Questions

The tangential force method involves finding the component of the applied force that is perpendicular to the displacement from the pivot point to where the force is applied. This perpendicular component of the force is called theĀ tangential force.
20061122120126329809921253700004765.png
(a) What isĀ F_t, the magnitude of the tangential force that acts on the pole due to the tension in the rope?

Express your answer in terms ofĀ TĀ andĀ theta.

When using the tangential force method, you calculate the torque using the equation

\tau = f_{\rm t} d,

whereĀ dĀ is the distance from the pivot to the point where the force is applied. The sign of the torque can be determined by checking which direction the tangential force would tend to cause the pole to rotate (where counterclockwise rotation implies positive torque).

(b) What is the magnitude of the torqueĀ tauĀ on the pole, about point A, due to the tension in the rope?
Express your answer in terms ofT,Ā L, andĀ theta.

The moment arm method involves finding the effective moment arm of the force. To do this, imagine a line parallel to the force, running through the point at which the force is applied, and extending off to infinity in either direction. You may shift the force vector anywhere you like along this line without changing the torque, provided you do not change the direction of the force vector as you shift it. It is generally most convenient to shift the force vector to a point where the displacement from it to the desired pivot point is perpendicular to its direction. This displacement is called theĀ moment arm.

For example, consider the force due to tension acting on the pole. Shift the force vector to the left, so that it acts at a point directly above point A in the figure. The moment arm of the force is the distance between the pivot and the tail of the shifted force vector. The magnitude of the torque about the pivot is the product of the moment arm and force, and the sign of the torque is again determined by the sense of the rotation of the pole it would cause.
20061122121536329809931322450004365.png
(c) FindĀ R_m, the length of the moment arm of the force.
Express your answer in terms ofLĀ andĀ theta.
2006112212386329809938811512508683.png

(d)Now consider a woman standing on the ball of her foot as shown. A normal force of magnitudeĀ NĀ acts upward on the ball of her foot. Achilles' tendon is attached to the back of the foot. The tendon pulls on the small bone in the rear of the foot with a forceF. This small bone has a lengthĀ x, and the angle between this bone and the Achilles' tendon isĀ phi. The horizontal displacement between the ball of the foot and the point P isĀ D.
D oll
Find Ļ„p, the torque about point P due to the force applied by the Achilles' tendon. Express your answer in terms of F, Ļ•, and x.

The tangential force method involves finding the component of the applied force that is perpendicular to the displacement from the pivot point to where the force is applied. This perpendicular component of the force is called the tangential force.
20061122120126329809921253700004765.png
(a) What isĀ F_t, the magnitude of the tangential force that acts on the pole due to the tension in the rope?

Express your answer in terms of TĀ andĀ theta.

When using the tangential force method, you calculate the torque using the equation

\tau = f_{\rm t} d,

whereĀ d is the distance from the pivot to the point where the force is applied. The sign of the torque can be determined by checking which direction the tangential force would tend to cause the pole to rotate (where counterclockwise rotation implies positive torque).

(b) What is the magnitude of the torqueĀ tau on the pole, about point A, due to the tension in the rope?
Express your answer in terms ofT,Ā L, andĀ theta.

The moment arm method involves finding the effective moment arm of the force. To do this, imagine a line parallel to the force, running through the point at which the force is applied, and extending off to infinity in either direction. You may shift the force vector anywhere you like along this line without changing the torque, provided you do not change the direction of the force vector as you shift it. It is generally most convenient to shift the force vector to a point where the displacement from it to the desired pivot point is perpendicular to its direction. This displacement is called the moment arm.

For example, consider the force due to tension acting on the pole. Shift the force vector to the left, so that it acts at a point directly above point A in the figure. The moment arm of the force is the distance between the pivot and the tail of the shifted force vector. The magnitude of the torque about the pivot is the product of the moment arm and force, and the sign of the torque is again determined by the sense of the rotation of the pole it would cause.
20061122121536329809931322450004365.png
(c) FindĀ R_m, the length of the moment arm of the force.
Express your answer in terms ofLĀ andĀ theta.
2006112212386329809938811512508683.png

(d)Now consider a woman standing on the ball of her foot as shown. A normal force of magnitude N acts upward on the ball of her foot. Achilles' tendon is attached to the back of the foot. The tendon pulls on the small bone in the rear of the foot with a forceF. This small bone has a lengthĀ x, and the angle between this bone and the Achilles' tendon is phi. The horizontal displacement between the ball of the foot and the point P is D.
D oll
Find Ļ„p, the torque about point P due to the force applied by the Achilles' tendon. Express your answer in terms of F, Ļ•, and x.

I would appreciate help with any of the following problems. Thankyou.

22. In an isometric exercise a person places a hand on a scale andpushes vertically downward, keeping the forearm horizontal. This ispossible because the triceps muscle applies an upward forceperpendicular to the arm, as the figure indicates. The forearmweighs 22.1 N and has a center of gravity as indicated. The scaleregisters 113 N. Determine the magnitude of .
Distance between the force applied by the muscle and the elbowjoint 0.025 m. The distance between the elbow joint and the armscenter of gravity is 0.15 m. Distance between the arms center ofgravity and the center of the scale is 0.15 m. I was not able tofind the answer to this problem using the solution posted on thewebsite.

26.A wrecking ball (weight = 4310 N) is supported by a boom, whichmay be assumed to be uniform and has a weight of 2550 N. As thedrawing shows, a support cable runs from the top of the boom to thetractor. The angle between the support cable and the horizontal is32ƂĀ°, and the angle between the boom and the horizontal is 48ƂĀ°. Find(a) the tension in the support cable and (b) the magnitude of theforce exerted on the lower end of the boom by the hinge at pointP.

56. Two identical wheels are moving on horizontal surfaces. Thecenter of mass of each has the same linear speed. However, onewheel is rolling, while the other is sliding on a frictionlesssurface without rolling. Each wheel then encounters an inclineplane. One continues to roll up the incline, while the othercontinues to slide up. Eventually they come to a momentary halt,because the gravitational force slows them down. Each wheel is adisk of mass 3.64 kg. On the horizontal surfaces the center of massof each wheel moves with a linear speed of 5.26 m/s. (a),(b) Whatis the total kinetic energy of each wheel? (c), (d) Determine themaximum height reached by each wheel as it moves up the incline. Iwas able to solve for the rolling wheel but couldn't find the rightanswers for the sliding wheel.

66. A thin uniform rod is rotating at an angular velocity of 1.89rad/s about an axis that is perpendicular to the rod at its center.As the figure indicates, the rod is hinged at two places,one-quarter of the length from each end. Without the aid ofexternal torques, the rod suddenly assumes a "u" shape, with thearms of the "u" parallel to the rotation axis. What is the angularvelocity of the rotating "u"?

76. A woman who weighs 501 N is leaning against a smooth verticalwall, as the drawing shows. Find (a) the magnitude of the force N(directed perpendicular to the wall) exerted on her shoulders bythe wall and the (b) horizontal and (c) vertical components of theforce exerted on her shoes by the ground.