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/s2and an angular acceleration of 1.2 rad/s2?

Answer 1

We can start by using the formula for centripetal acceleration:

a = v^2/r

where a is the centripetal acceleration, v is the velocity of the object, and r is the radius of the circular path.

To find v, we can use the formula for tangential acceleration:

a_t = rα

where a_t is the tangential acceleration and α is the angular acceleration.

Rearranging this formula, we get:

α = a_t/r

v = a_t/α

Substituting the given values, we get:

α = 1.2 rad/s^2
a_t = 3.5 m/s^2
r = 1.5 m

v = a_t/α = 3.5 m/s^2 / 1.2 rad/s^2 = 2.92 m/s

Finally, we can use the formula for centripetal acceleration to find a:

a = v^2/r = (2.92 m/s)^2 / 1.5 m = 5.69 m/s^2

Therefore, the centripetal acceleration of the object is 5.69 m/s^2.