For a particle in uniform circular motion, the acceleration a at any point P(R, θ) on the circular path of radius R is (when θ is measured from the positive x-axis and v is uniform speed):
To solve the problem of finding the acceleration of a particle in uniform circular motion at a point \(P(R, \theta)\) on the circular path, we need to understand the concepts of circular motion and acceleration.
In uniform circular motion, a particle moves in a circle at a constant speed. However, its velocity is constantly changing direction, leading to an acceleration towards the center of the circle, known as centripetal acceleration.
The centripetal acceleration \(a\) of a particle in uniform circular motion is given by the formula:
a = \frac{v^2}{R}
Where:
v is the uniform speed of the particle.
R is the radius of the circular path.
The direction of this acceleration is always towards the center of the circle. In terms of vector components, we can express this in terms of the angles given (with respect to the x-axis). At point \(P(R, \theta)\), the radial direction can be given by the unit vector which points inward, towards the center of the circle:
The particle is at an angle \(\theta\) from the positive x-axis. Therefore, the components of the acceleration vector in terms of i (x-direction) and j (y-direction) are:
In the x-direction (i component): -\frac{v^2}{R}\cos\theta. The negative sign indicates that the acceleration is directed towards the center of the circle, against the outward (positive) radial direction.
In the y-direction (j component): -\frac{v^2}{R}\sin\theta. Similarly, the negative sign ensures that the acceleration is directed towards the center.
Thus, the total acceleration vector a can be expressed as:
a = -\frac{v^2}{R}\cos\theta\, \mathbf{i} - \frac{v^2}{R}\sin\theta\, \mathbf{j}
After examining the given options to reason about which one matches the derived result, we find:
