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Physics LibreTexts

8.5: In Summary

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  1. To solve problems involving motion in two dimensions, you should break up all the forces into their components along a suitable pair of orthogonal axes, then apply Newton’s second law to each direction separately: Fnet,x=max, Fnet,y=may. It is convenient to choose your axes so that at least one of either ax or ay will be zero.
  2. An object thrown with some horizontal velocity component and moving under the influence of gravity alone (near the surface of the Earth) will follow a parabola in a vertical plane. This results from horizontal motion with constant velocity, and vertical motion with constant acceleration equal to g, as described by equations (8.2.4).
  3. To analyze motion up or down an inclined plane, it is convenient to choose your axes so that the x axis lies along the surface, and the y axis is perpendicular to the surface. Then, if θ is the angle the incline makes with the horizontal, the force of gravity on the object will also make an angle θ with the negative y axis.
  4. Recall that the force of kinetic friction will always point in a direction opposite the motion, and will have magnitude Fk=μkFn, whereas the force of static friction will always take on whatever value is necessary to keep the object from moving, up to a maximum value of Fsmax=μsFn.
  5. An object moving in an arc of a circle of radius R with a speed v experiences a centripetal acceleration of magnitude ac=v2/R. “Centripetal” means the corresponding vector points towards the center of the circle. Accordingly, to get an object of mass m to move on such a path requires a centripetal force Fc=mv2/R.
  6. To describe the motion of a particle on a circle of radius R, we use an angular position variable θ(t), in terms of which we define angular displacement Δθ, angular velocity ω=dθ/dt, and angular acceleration α=dω/dt. The equations for motion in one dimension with constant acceleration apply to circular motion with constant α with the changes xθ, vω and aα.
  7. The displacement along the circle, s, corresponding to an angular displacement Δθ, is (in magnitude) s=R|Δθ|. Similarly, the (linear) speed of the particle (magnitude of its velocity vector) is equal to v=R|ω|, and the tangential component of its acceleration vector has magnitude at=R|α|. In addition to this, the particle always has a radial acceleration component ar equal to the centripetal acceleration of point 5 above.

This page titled 8.5: In Summary is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Julio Gea-Banacloche (University of Arkansas Libraries) via source content that was edited to the style and standards of the LibreTexts platform.

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