Answer:
600.6 m/s^2
Explanation:
α = 155 rad/s^2
ω = 20 rad/s
r = 1.4 m
Tangential acceleration, aT = r x α = 1.4 x 155 = 217 m/s^2
Centripetal acceleration, ac = rω^2 = 1.4 x 20 x 20 = 560 m/s^2
The tangential acceleration and the centripetal acceleration both are perpendicular to each other. Let a be the resultant acceleration.
a^2 = aT^2 + ac^2
a^2 = 217^2 + 560^2
a = 600.6 m/s^2
The magnitude of the total acceleration of the top of the racket is 600.6 m/s².
The given parameters;
The centripetal acceleration of the racket is calculated as follows;
[tex]a_c = \omega ^2 r\\\\a_c = (20)^2 \times 1.4\\\\a_c = 560 \ m/s^2[/tex]
The tangential acceleration is calculated as;
[tex]a_t = \alpha _r \times r\\\\a_t = 155 \times 1.4\\\\a_t = 217 \ m/s^2[/tex]
The magnitude of the total acceleration of the top of the racket is calculated as follows;
[tex]a= \sqrt{a_t ^2 + a_c^2} \\\\a = \sqrt{217^2 + 560^2} \\\\a = 600.6 \ m/s^2[/tex]
Thus, the magnitude of the total acceleration of the top of the racket is 600.6 m/s².
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