Question

A particle moves along a circle with radius R, so that the tangential component of its acceleration is constant. At t = 0 the velocity of the particle was equal to zero. Find

(a) the normal component an of the acceleration as function of time.
(b) the magnitude of the acceleration vector a as well as the angle the vector a forms with the position vector r as functions of time.

Answer 1

To find the normal component of acceleration, differentiate the velocity and acceleration functions.

Step-by-step explanation:

In order to find the normal component of the acceleration, we need to first find the velocity and acceleration vectors as functions of time. Using the given position function, we can differentiate it to find the velocity function:

v(t) = dr/dt = -3(4sin(3t)i + 4cos(3t)ĵ)

Next, we can differentiate the velocity function to find the acceleration function:

a(t) = dv/dt = -3(3cos(3t)i - 3sin(3t)ĵ)

Since the tangential component of the acceleration is constant, we can set the coefficient of the tangential component equal to a constant value:

-3(3sin(3t)) = Constant

Solving this equation will give us the value of the constant. Once we know the value of the constant, we can substitute it back into the acceleration function to find the normal component of the acceleration as a function of time.