Answer:
The change in the tire’s angular velocity is -3.4rad/s and the tire’s average angular acceleration is [tex]2.72rad/s^2.[/tex]
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:
[tex]\Delta \omega = \omega_f - \omega_i[/tex]
Where \omega means the angular velocity.
Our values are given by:
[tex]\omega_i = 3.4rad/s[/tex]
[tex]\omega_f = -3.4rad/s[/tex]
There was a change in the direction of the speed.
Then the Total change of velocity is
[tex]\Delta omega = -3.4-3.4 = -6.8rad/s[/tex]
We can know find the Acceleration of the object, which is given by,
[tex]\alpha = \frac{\omega}{t}[/tex]
[tex]\alpha = \frac{-6.8rad/s}{2.5}[/tex]
[tex]\alpha = 2.72rad/s^2[/tex]
Therefore the change in the tire’s angular velocity is [tex]-3.4rad/s[/tex] and the tire’s average angular acceleration is [tex]2.72rad/s^2.[/tex]
Answer:
The change in the tire's angular velocity is [tex]6.80\frac{rad}{s}[/tex]. The tire's average angular acceleration is [tex]2.72\frac{rad}{s^2}[/tex]
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
[tex]\Delta\omega=\omega_{f}-\omega_{i}=3.40\frac{rad}{s}-(-3.40\frac{rad}{s})=6.80\frac{rad}{s}[/tex]
On the other hand, to calculate the average angular acceleration we have that
[tex]\alpha_{av}=\frac{\Delta\omega}{\Delta t}[/tex]
we just calculated Δω, and Δt is given in the problem, therefore
[tex]\alpha_{av}=\frac{\Delta\omega}{\Delta t}=\frac{6.80rad}{2.50s^2}=2.72\frac{rad}{s^2}[/tex]
is the average angular acceleration of the tire.
