Angular Motion

However the farther a particle is from the axis the

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Unformatted text preview: same amount of time; that is, they all have the same angular speed ω. However, the farther a particle is from the axis, the greater the circumference of its circle and so the faster its linear speed v must be. You can notice this if you are on a merry-go-round. You turn with the same angular speed ω regardless of your distance from the center, but your linear speed v increases noticeably if you move to the outside edge of the merry-to-round. We often relate the linear variables s, v, and a for a particular point in a rotating body to the angular Figure 4 View of a particle moving along a variables θ, ω, and α for that body. The two sets of circular path variables are related by r, the perpendicular distance of the point from the axis of rotation axis. This distance is a straight line between the point and the axis of rotation. It is also the radius of the circle traveled by the point around the axis of rotation. 2.7.1 The Position If a reference line on a rigid body rotates through an angle θ, a point on the reference line at a position r from the axis of rotation (See Figure 4) moves a distance s along a circular arc which can be defined as (12) This is our linear-angular relation. CAUTION: The angle θ here must be in radians because equation (12) is itself the definition of angular measure in radians. See Equation (2). 2.7.2 The Speed Now suppose that we want to find the change in distance s for a constant r. Thus we can say the change in s is ∆ ∆ (13) Now let us suppose that we want to find out how the distance changes in some time interval Δt. In order to do this, we must divide equation (13) by Δt which yields ∆ ∆ ∆ ∆ (14) (15) 5 If we assume that Δt is very small, then we can say that we are have instantaneous velocity. Thus equation (14) becomes (16) This equation shows that the linear speed of a point on a rotating object increases as that point is moved outward from the center of rotation toward the rim; however, every point on the rotating object has the same angular speed. 2.7.3 The Acceleration Now let us suppose that the particle in Figure 4 has a changing angular speed Δω in the time interval Δt. Thus using equation (16), we see that ∆ ∆ ∆ ∆ (17) a (18) If we assume that the time interval Δt is very small then equation (18) becomes a (19) Thus the tangential acceleration of a point on a rotating object equals the distance of that point from the axis of...
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This document was uploaded on 03/20/2014 for the course PHYS 215 at Lafayette.

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