Answer:
a) [tex] a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}[/tex]
b) [tex] k = \frac{9.8}{9.74}=1.006[/tex]
Explanation:
Part a
For this case we can begin finding the period like this:
[tex] T= \frac{1}{w} =\frac{1}{617 rad/s}=0.00162 s[/tex]
Then we know that the centripetal acceleration is given by:
[tex] a= \frac{v^2}{r}[/tex]
And the velocity is given by:
[tex] v=\frac{2\pi r}{T}[/tex]
If we replace this into the acceleration we got:
[tex] a = \frac{(\frac{2\pi r}{T})^2}{r}= \frac{4 \pi^2 r}{T^2}[/tex]
And we can replace the values and we got:
[tex] a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}[/tex]
Part b
For this case we want to find a value of k such that:
[tex] a= k 9.8[/tex]
Where a = 9.74, so then we can solve for k like this:
[tex] k = \frac{9.8}{9.74}=1.006[/tex]
a) The magnitude of the centripetal acceleration be "300581.71 m/s²".
b) The acceleration be "30671.5 g"
According to the question,
Radius, r = 0.020m
g = 9.8 m/s²
[tex]\omega[/tex] = 617 rev/s
By converting it into "rads/sec",
= 617 × [tex]\frac{2 \pi}{1 \ rev}[/tex]
= 3876.73 rads/sec
We know the formula,
(a)
Centripetal acceleration, [tex]a_c[/tex] = [tex]\frac{v^2}{r}[/tex]
= r[tex]\omega^2[/tex]
By substituting the values, we get
= 0.020 × 3876.73
= 300581.71 m/s²
(b)
As a multiple of "g", the acceleration be:
[tex]a_c[/tex] = 30671.5 g
Thus the answer above is appropriate.
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