Answer:
Part(a): The angular acceleration is [tex]\bf{1410.78~rad/s^{2}}[/tex].
Part(b): The linear acceleration is [tex]\bf{28.21~m/s^{2}}[/tex].
Explanation:
Given:
The angular velocity, [tex]\omega = 235.6~rad/s[/tex]
Time taken, [tex]t = 0.167~s[/tex]
Diameter of the disk, [tex]d = 0.12~m[/tex]
Radius of the concerned point, [tex]r_{p} = \dfrac{1}{3}(d/2)[/tex].
Part(a):
The angular acceleration ([tex]\alpha[/tex]) is given by
[tex]\alpha &=& \dfrac{\omega}{t}\\&=& \dfrac{235.6~rad/s}{0.167~s}\\&=& 1410.78~rad/s^{2}[/tex]
Part(b):
The radius ([tex]r_{p}[/tex]) of the concerned point is given by
[tex]r_{p} &=& \dfrac{1}{3}\dfrac{0.12~m}{2}\\&=& 0.02~m[/tex]
Linear acceleration ([tex]a[/tex]) is give by
[tex]a &=& \alpha \times r_{p}\\&=& (1410.78~rad/sec)(0.02~m)\\&=& 28.21~m/s^{2}[/tex]
This question involves the concepts of equations of motion for angular motion and tangential acceleration.
a) The magnitude of the average angular acceleration is "1410.78 rad/s²".
b) The magnitude of the tangential acceleration of a point 1/3 of the way out from the center of the disk is "28.22 m/s²".
a)
We will use the first equation of motion for the angular motion to find out the angular acceleration:
[tex]\alpha = \frac{\omega_f-\omega_i}{t}[/tex]
where,
[tex]\alpha[/tex] = angular acceleration = ?
[tex]\omega _f[/tex] = final angular speed = 0 rad/s
[tex]\omega_i[/tex] = initial angular acceleration = 235.6 rad/s
t = time = 0.167 s
Therefore,
[tex]\alpha=\frac{0\ rad/s-235.6\ rad/s}{0.167\ s}\\\\\alpha = 1410.78\ rad/s^2[/tex]
b)
The tangential acceleration can be found using the following formula:
[tex]a_T=(r)(\alpha)[/tex]
where,
[tex]a_T[/tex] = tangential acceleration = ?
r = distance of point from center = [tex]\frac{1}{3}(\frac{d}{2}) = \frac{1}{3}(\frac{0.12\ m}{2}) = 0.02\ m[/tex]
Therefore,
[tex]a_T=(0.02\ m)(1410.78\ rad/s^2)\\a_T=28.22\ m/s^2[/tex]
Learn more about the angular motion here:
brainly.com/question/14979994?referrer=searchResults
The attached picture shows the angular equations of motion.
