Question

A bicycle tire is spinning counterclockwise at 2.30 rad/s. During a time period Δt = 2.20 s, the tire is stopped and spun in the opposite (clockwise) direction, also at 2.30 rad/s. Calculate the change in the tire's angular velocity Δ and the tire's average angular acceleration av. (Indicate the direction with the signs of your answers.)

(a)

the change in the tire's angular velocity Δ (in rad/s)

rad/s

(b)

the tire's average angular acceleration av (in rad/s2)

rad/s

Answer 1

The change in the tire's angular velocity Δ is -6.6.0 rad/s and he tire's average angular acceleration av is -4.40 rad/s^2

To calculate the change in the tire's angular velocity Δ, we need to determine the difference between the initial and final angular velocities.

In this case, the initial angular velocity is 3.30 rad/s in the counterclockwise direction, and the final angular velocity is also 3.30 rad/s, but in the opposite (clockwise) direction.

Since the angular velocity is a vector quantity, we need to consider the signs. Counterclockwise is considered positive, while clockwise is considered negative.

The initial angular velocity is +3.30 rad/s, and the final angular velocity is -3.30 rad/s.

Therefore, the change in the tire's angular velocity Δ is given by:

Δ = final angular velocity - initial angular velocity

= (-3.30 rad/s) - (+3.30 rad/s)

= -6.60 rad/s

So, the change in the tire's angular velocity Δ is -6.60 rad/s.

To calculate the tire's average angular acceleration av, we can use the formula:

av = Δangular velocity/time interval

Since we already know the change in the tire's angular velocity Δ is -6.60 rad/s and the time interval Δt is 1.50 s, we can substitute these values into the formula:

av = (-6.60 rad/s) / (1.50 s)

= -4.40 rad/s^2

Therefore, the tire's average angular acceleration av is -4.40 rad/s^2.

In summary:

(a) The change in the tire's angular velocity Δ is -6.60 rad/s.

(b) The tire's average angular acceleration av is -4.40 rad/s^2.