Respuesta :
Answer:
110.9 m/s²
Explanation:
Given:
Distance of the tack from the rotational axis (r) = 37.7 cm
Constant rate of rotation (N) = 2.73 revolutions per second
Now, we know that,
1 revolution = [tex]2\pi[/tex] radians
So, 2.73 revolutions = [tex]2.73\times 2\pi=17.153\ radians[/tex]
Therefore, the angular velocity of the tack is, [tex]\omega=17.153\ rad/s[/tex]
Now, radial acceleration of the tack is given as:
[tex]a_r=\omega^2 r[/tex]
Plug in the given values and solve for [tex]a_r[/tex]. This gives,
[tex]a_r=(17.153\ rad/s)^2\times 37.7\ cm\\a_r=294.225\times 37.7\ cm/s^2\\a_r=11092.28\ cm/s^2\\a_r=110.9\ m/s^2\ \ \ \ \ \ \ [1\ cm = 0.01\ m][/tex]
Therefore, the radial acceleration of the tack is 110.9 m/s².
The required value of the radial acceleration of a tack stuck in the tire is [tex]110.9 \;\rm m/s^{2}[/tex].
Radial Acceleration:
An object moving along the circular path such that the rate of change of angular velocity, directed towards the circumferential center, is known as radial acceleration.
Given data:
The radial distance from the axis is, r = 37.7 cm.
The rate of rotation is, N = 2.73 rotations per second (rps)
The expression for the radial acceleration is given as,
[tex]a_{r}= \omega^{2} \times r[/tex]
here, [tex]\omega[/tex] is the angular velocity.
Since, 1 revolution = 2π radians.
Then angular velocity (in rad/s) is calculated as,
[tex]\omega = N \times 2 \pi\\\\ \omega = 2.73 \times 2 \pi\\\\ \omega =17.153 \;\rm rad/s[/tex]
Then the radial acceleration is calculated as,
[tex]a_{r} = (17.153)^{2} \times 37.7\\\\ a_{r}=11092.28 \;\rm cm/s^{2}[/tex]
Converting into m/s² as,
[tex]a_{r} = 11092.28 \times 0.01 \\\\ a_{r}=110.9 \;\rm m/s^{2}[/tex]
Thus, we can conclude that the required value of the radial acceleration of a tack stuck in the tire is [tex]110.9 \;\rm m/s^{2}[/tex].
Learn more about the radial acceleration here:
https://brainly.com/question/13943269
