Question

A 2.00 kg object is moving in a circular path with a radius of 5.00 cm. The object starts from rest and with constant angular acceleration, obtains an angular velocity of 6.00 rad/s in 3.00 s. The object then comes to a stop with constant deceleration in 4.00 s. The centripetal component of acceleration of the object at 2.00 s is​

Answer 1

Answer:

0.800 m/s²

Explanation:

First, calculate the angular acceleration:

ω = αt + ω₀

6.00 rad/s = α (3.00 s) + 0 rad/s

α = 2.00 rad/s²

Now calculate the angular velocity at t = 2.00 s:

ω = αt + ω₀

ω = (2.00 rad/s²) (2.00 s) + 0 rad/s

ω = 4.00 rad/s

Calculate the linear velocity:

v = ωr

v = (4.00 rad/s) (0.0500 m)

v = 0.200 m/s

Finally, calculate the centripetal acceleration:

a = v² / r

a = (0.200 m/s)² / (0.0500 m)

a = 0.800 m/s²