What is the centripetal acceleration of an object moving on a circular path of 1.5 m if it has an acceleration of 3.5 m/s2 and an angular acceleration of 1.2

Question

rad/s2?
A. 3.0 rad/s2
B. 3.3 rad/s2
C. 3.7 rad/s2
D. 4.5 rad/s2

Answer 1

The centripetal acceleration (a) of an object moving on a circular path can be calculated using the formula:

a = r * ω^2

where r is the radius of the circular path and ω is the angular velocity (ω = Δθ/Δt).

In this question, we are given the radius of the circular path (r = 1.5 m) and the angular acceleration (α = 1.2 rad/s^2). To find the angular velocity, we can use the formula:

α = dω/dt

where dω is the change in angular velocity and dt is the change in time. Since the object is moving on a circular path with constant speed, there is no change in linear velocity and hence no change in angular velocity. Therefore, ω is constant.

We are also given the acceleration (a) of the object, which is equal to the centripetal acceleration (since there is no tangential acceleration in uniform circular motion). Thus, we have:

a = r * ω^2

3.5 m/s^2 = 1.5 m * ω^2

ω^2 = 2.333...

ω = √2.333... = 1.527 rad/s

Therefore, the centripetal acceleration is:

a = r * ω^2 = 1.5 m * (1.527 rad/s)^2 = 3.5 m/s^2

The answer is A. 3.0 rad/s^2.