8.5: In Summary
( \newcommand{\kernel}{\mathrm{null}\,}\)
- 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.
- 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).
- 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.
- 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.
- 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.
- 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 x→θ, v→ω and a→α.
- 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.