# EQUILIBRIO ROTACIONAL Y TRASLACIONAL PDF

F2=()=8N URL del artículo: Fuente: Ejemplos de Equilibrio rotacional y traslacional. Quiz TORQUE. 9º Secundaria, torque, equilibrio rotacional, momento de fuerza o torque. Ferney Rosero Hernández. TORQUE O MOMENTO DE FUERZA. Ejemplos de equilibrio traslacional. Diana Rueda. Equilibrio rotacional. Moisés Galarza Espinoza. ¿Qué es Threat Hunting y por qué lo. Author: Mutaxe Gorg Country: Chad Language: English (Spanish) Genre: Relationship Published (Last): 18 March 2013 Pages: 405 PDF File Size: 11.14 Mb ePub File Size: 10.75 Mb ISBN: 303-9-12145-322-2 Downloads: 1197 Price: Free* [*Free Regsitration Required] Uploader: Meztilabar ### Inercia Rotacional by Sebastian Zelaya Paez on Prezi

Published on Apr View 7. The bicycle was so named because the size relationship of its two wheels was about the same as the size relationship of the traslaciinal and the farthing, two English coins. When the rider looks down at the top of the front wheel, he sees it moving forward faster than he and the handlebars are moving. Yet the center of the wheel does not appear to be moving at all relative to the handlebars. How can different parts of the rolling wheel move at different linear speeds? We begin by describing such motion, which is called rolling motion.

The central topic of this chapter is, however, angular momentum, a quantity that plays a key role in rotational dynamics. In analogy to the conservation of linear momentum, we nd that the angular momentum of a rigid object is always conserved if no external torques act on the object. Like the law of conservation of linear momentum, the law of conservation of angular momentum is a fundamental law of physics, equally valid for relativistic and quantum systems.

In general, such motion is very complex. However, we can simplify matters by restricting our discussion to a homogeneous rigid object having a high degree of symmetry, such as a cylinder, sphere, or hoop.

Furthermore, we assume that the trasacional undergoes rolling motion along a at surface. We shall see that if an object such as a cylinder rolls without slipping on the surface we call this pure rolling motiona simple relationship exists between its rotational and translational motions. Suppose a cylinder is rolling on a straight path. This means that the axis of rotation remains parallel to its initial orientation in space. Consider a uniform cylinder of radius R rolling without slipping on a horizontal surface Fig.

As the cylinder equilihrio through an angleits center of mass moves a linear distance s R see Eq. Therefore, the linear speed of the center of mass for pure rolling motion is given by v CM ds dt R d dt R The center moves traslcaional a straight line green linewhereas the point on the rim moves in the path called a cycloid red curve.

HenryLeap and Jim Lehman The linear velocities of the center of mass and of various points on and within the cylinder are illustrated in Figure A short time after the moment shown in the drawing, the rim point labeled P will have rotated from the six oclock position to, say, the seven oclock position, the point Q will have rotated from the ten oclock position to the eleven oclock position, and so on.

Note that the linear velocity of any point is in a direction perpendicular to the line from that point to the contact point P. At any instant, the part of the rim that is at point P is at rest relative to the surface because slipping does not occur. All points on the cylinder have the same angular speed.

To see why this is so, let us model the rolling motion of the tarslacional in Figure For the pure translational motion shown in Figure For the pure rotational motion shown in Figure The combination of these two motions represents the rolling motion shown in Figure Note in Figure As noted earlier, the center of mass moves with linear speed v CM while the contact point between the surface and cylinder has a linear speed of zero. We can express the total kinetic energy of the rolling cylinder as7.

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In other words, all points rotate about P. The center of mass of the object moves with a traslaciobal vCMand the point P moves with a velocity 2vCM.

K1 2I P2 Mechanical energy is conserved if no slipping occurs. The term 1 I CM 2 represents the rotational kinetic energy of the cylinder about its 2 center of mass, and the term 1Mv CM2 represents the kinetic energy the cylinder 2 would have if it were just translating through space without rotating.

## 11-Movimiento Rotacional y Momento Angular

Thus, we can say that the total kinetic energy of a rolling object is the sum of the rotational kinetic energy about the center of mass and the translational kinetic rotacionnal of the center of mass. We can use energy methods to treat a class of problems concerning the rolling motion of a sphere down a rough incline. The analysis that follows also applies to the rolling motion of a cylinder or hoop. We assume that the sphere in Figure Note that accelerated rolling motion is possible only if a frictional force is present between the sphere and the incline to produce a net torque about the center of mass.

Despite the presence of friction, no loss of mechanical energy occurs because the contact point is at rest relative to the surface at any instant. On the other hand, if the sphere were to slip, mechanical energy would be lost as motion progressed. Using the fact that v CM R for pure rolling motion, we can traskacional Equation Because the sphere starts from rest at the top, its kinetic traslacionql at the bottom, given by Equation Therefore, the speed of the center of mass at the bottom can be obtained by equating these two quantities: It quickly stops moving because of friction between it and the oor.

Yet, if you were to start a basketball rolling with the same initial speed, it would probably keep rolling from one end of the gym to the other. Why does a basketball roll so far? Doesnt friction affect its motion? Hence, after squaring both sides, we can express the equation above as v CM2 10 gx sin 7 Comparing this with the expression from kinematics, v CM2 equilubrio CMx see Eq. The sphere starts from the top of the incline with potential energy U g Mgh and kinetic energy K 0.

As we equiljbrio seen before, if it fell vertically from that height, it would have a linear speed of! After rolling down the incline, the linear speed of the center of mass must be less than rotaciinal value because some of the initial traslacipnal energy is diverted into rotational kinetic energy rather than all being converted into rotaxional kinetic energy. To calculate the linear acceleration of the center of mass, we note that the vertical displacement is related to the displacement x along the incline through the relationship hThese results are quite interesting in that both the speed and the acceleration of the center of mass are independent of the mass and the radius of the sphere!

That is, all dquilibrio solid spheres experience the same speed and acceleration on a given incline. If we repeated traslacoinal calculations for a hollow sphere, a solid cylinder, or a hoop, we would obtain similar results in which only the factor in front of g sin would differ.

The constant factors that appear in the expressions for v CM and a CM depend only on the moment of inertia traslacionnal the center of mass for the specic body.

In all cases, the acceleration of the center of mass is less than g sinthe value the acceleration would have if the incline were rotacionall and no rolling occurred. The free-body diagram for the sphere is illustrated in Figure Solutionmass givesNewtons second law applied to the center ofwhere x is measured along the slanted surface equipibrio the incline. Now let us write an expression for the torque acting on the sphere.

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Free-body diagram for a solid sphere rolling down anthrough the center of the sphere and is perpendicular to the plane of the gure. However, the force of static friction produces a torque about this axis equal to fR in the clockwise direction; therefore, because is also in thewhich agrees with the result of Example Note that F ma applies only if F is the net external force exerted on the sphere and a is the acceleration of its center of mass.

In the case of our sphere rolling down an equioibrio, even though the frictional trxslacional does not change the total kinetic energy of the sphere, it does contribute to F and thus decreases the acceleration of the center of mass. As a result, the nal translational kinetic energy is less than it would be in the absence of friction.

As mentioned in Example QuickLabHold a basketball and a tennis ball side by side at the top of a ramp and release them at the same time. Which reaches the bottom rst? Does the outcome depend on the angle of the ramp? What if the equillibrio were 90 eotacional is, if the balls were in free fall? Equiljbrio origin O is assumed to be in an inertial frame, so Newtons rst law is valid in this case.

As we saw in Section The axis about which F tends to produce rotation is perpendicular to the plane formed by r and F. If the force lies in the xy plane, as it does in Figure The force in Figure If we reversed the direction of F in Figure The torque involves the two vectors r and F, and its direction is perpendicular to the plane of r and F. We can establish a mathematical relationship betweenr, and F, using a new mathematical operation called the vector product, or cross product: The direction of C is perpendicular to the plane formed by A and B, and the best way to determine this direction is to use the right-hand rule illustrated in Figure The four ngers of the right hand are pointed along A and then wrapped into B through the angle.

The direction of the erect right thumb is the direction of A B Equilibgio. rotaional Because of the notation, A B is often read A cross B; hence, the term cross product. Some properties of the vector product that equilibbrio from its denition are as follows: Unlike the scalar product, the vector product is not commutative.

Instead, the order in which the two vectors are multiplied in a cross product is important: Therefore, if you change the order of the vectors in a cross product, dquilibrio must change the sign. You could easily verify this relationship with the right-hand rule. The vector product obeys the distributive law: Traslacionql is left as an exercise to show from Equations The direction of C is perpendicular to the plane formed by A and B, and this direction is determined by the righthand rule.

Cross products of unit vectorsi i j ki j k ijj j k ik i j kk k i j0 The cross product of any two vectors A and B can be expressed in the following determinant form: Solutionobtain A BUsing Equations As an alternative method for nding A B, we could use Equation A B 0 i 0 j [ 2 2 3 1 ]k 7k i We have omitted the terms containing i equlibrio and j j because, as Equation A skater glides rapidly toward the pole, aiming a little to the side so that she does not hit it.