A particle moves along circular path with constant tangential acceleration of 0.28 m/s2. It starts at rest from point A. Find velocity and acceleration component of the particle along x and y direction, when it reaches point B.
Answers
Explanation:
In one-dimensional kinematics, objects with a constant speed have zero acceleration. However, in two- and three-dimensional kinematics, even if the speed is a constant, a particle can have acceleration if it moves along a curved trajectory such as a circle. In this case the velocity vector is changing, or
This is shown in (Figure). As the particle moves counterclockwise in time
on the circular path, its position vector moves from
.
The velocity vector has constant magnitude and is tangent to the path as it changes from
,
changing its direction only. Since the velocity vector
is perpendicular to the position vector
the triangles formed by the position vectors and
and the velocity vectors and
are similar. Furthermore, since
the two triangles are isosceles. From these facts we can make the assertion
Figure a shows a circle with center at point C. We are shown radius r of t and radius r of t, which are an angle Delta theta apart, and the chord length delta r connecting the ends of the two radii. Vectors r of t, r of t plus delta t, and delta r form a triangle. At the tip of vector r of t, the velocity is shown as v of t and points up and to the right, tangent to the circle. . At the tip of vector r of t plus delta t, the velocity is shown as v of t plus delta t and points up and to the left, tangent to the circle. Figure b shows the vectors v of t and v of t plus delta t with their tails together, and the vector delta v from the tip of v of t to the tip of v of t plus delta t. These three vectors form a triangle. The angle between the v of t and v of t plus delta t is theta.
Figure 4.18 (a) A particle is moving in a circle at a constant speed, with position and velocity vectors at times
