Abicycle tire is spinning clockwise at 3.40 rad/s. during a time period δt = 2.50 s, the tire is stopped and spun in the opposite (counterclockwise) direction, also at 3.40 rad/s. calculate the change in the tire's angular velocity δω and the tire's average angular acceleration αav. (indicate the direction with the signs of your answers.)

The change in the tire’s angular velocity is -3.4rad/s and the tire’s average angular acceleration is

Explanation:

To solve this exercise we must go back to the kinematic equations of motion in which the angular velocity change is defined, and the expression derived from the expression from time and velocity. The equation of angular velocity change is given by:

Where \omega means the angular velocity.

Our values are given by:

There was a change in the direction of the speed.

Then the Total change of velocity is

We can know find the Acceleration of the object, which is given by,

Therefore the change in the tire’s angular velocity is and the tire’s average angular acceleration is

The change in the tire's angular velocity is . The tire's average angular acceleration is

Explanation:

Let's assume that the counterclockwise direction is the positive direction, then, as we were given the initial and final angular velocity (ω) and there is a direction change between them, we can calculate the change in angular velocity as

On the other hand, to calculate the average angular acceleration we have that

we just calculated Δω, and Δt is given in the problem, therefore

is the average angular acceleration of the tire.

i think the answer is a, hope this is right, that would be my best guess.

answer: good job it's all correct !

explanation: