Respuesta :
Answer:
0.38963 m/s²
1.246819325 m/s
0.38963 m/s²
0.95629 m/s²
1.03261 m/s² and 22.17°
Explanation:
d = Diameter of rim = 12.5 in = [tex]12.5\times 0.0254=0.3175\ m[/tex]
r = Radius = [tex]\frac{d}{2}=\frac{0.3175}{2}=0.15875\ m[/tex]
[tex]\omega_f[/tex] = Final angular velocity = [tex]75\times\frac{2\pi}{60}=7.85398\ rad/s[/tex]
[tex]\omega_i[/tex] = Initial angular velocity = 0
[tex]\alpha[/tex] = Angular acceleration
t = Time taken = 3.2 s
Equation of rotational motion
[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{7.85398-0}{3.2}\\\Rightarrow \alpha=2.45436875\ rad/s^2[/tex]
Tangential acceleration is given by
[tex]a_t=r\alpha\\\Rightarrow a_t=0.15875\times 2.45436875\\\Rightarrow a_t=0.38963\ m/s^2[/tex]
The tangential acceleration of the bug is 0.38963 m/s²
Tangential velocity is given by
[tex]v=r\omega\\\Rightarrow v=0.15875\times 7.85398\\\Rightarrow v=1.246819325\ m/s[/tex]
The tangential velocity of the bug is 1.246819325 m/s
The tangential acceleration is constant which is 0.38963 m/s²
Centripetal acceleration is given by
[tex]a_c=\frac{a_tt^2}{r}\\\Rightarrow a_c=\frac{0.38963^2\times 1}{0.15875}\\\Rightarrow a_c=0.95629\ m/s^2[/tex]
The centripetal acceleration of the bug is 0.95629 m/s²
The resultant of the acceleration gives us total acceleration
[tex]a=\sqrt{a_t^2+a_c^2}\\\Rightarrow a=\sqrt{0.38963^2+0.95629^2}\\\Rightarrow a=1.03261\ m/s^2[/tex]
Direction is given by
[tex]\theta=cos^{-1}\frac{a_c}{a}\\\Rightarrow \theta=cos^{-1}\frac{0.95629}{1.03261}\\\Rightarrow \theta=22.17^{\circ}[/tex]
The magnitude and direction of the acceleration is 1.03261 m/s² and 22.17°
a. The magnitude of the tangential acceleration of a bug on the rim is[tex]alpha = 2.45\;rad/s^2[/tex]
b. The tangential velocity of the bug, when the disk is at its final speed is 1.25 m/s.
c. The tangential acceleration, one second after the bug starts from rest is [tex]alpha_t = 0.39\; m/s^2[/tex]
d. The centripetal acceleration, one second after the bug starts from rest is [tex]alpha_c = 0.96 \;m/s^2[/tex]
e. The total acceleration, one second after the bug starts from rest is [tex]T_a = 1.04 \; m/s^2[/tex]
Given the following data:
- Diameter = 12.5 inches
- Initial angular velocity = 0 rev/min (since the disk is at rest).
- Final angular velocity = 75.0 rev/min
- Time = 3.20 seconds.
Conversion:
1 rev/min = 0.1047 rad/s
75.0 rev/min = X rad/s
Cross-multiplying, we have:
[tex]X = 75 \times 0.1047[/tex]
X = 7.85 rad/s
[tex]Radius = \frac{Diameter}{2} = \frac{12.5}{2} = 6.25\; in[/tex]
1 inch = 0.0254 meter
6.25 inches = 0.1588 meter
a. To find the magnitude of the tangential acceleration of a bug on the rim:
[tex]\alpha = \frac{w_f \; -\;w_i}{t} \\\\\alpha = \frac{ 7.85\; -\;0}{3.2}\\\\\alpha = 2.45\;rad/s^2[/tex]
b. To find the tangential velocity of the bug, when the disk is at its final speed:
[tex]V = rw\\\\V = 0.1588 \times 7.85[/tex]
V = 1.25 m/s
c. To find the tangential acceleration, one second after the bug starts from rest:
[tex]\alpha_t = r \alpha\\\\\alpha_t = 0.1588 \times 2.45\\\\ \alpha_t = 0.39\; m/s^2[/tex]
d. To find the centripetal acceleration, one second after the bug starts from rest:
[tex]\alpha_c = \frac{\alpha_t^2 t^2}{r} \\\\\alpha_c = \frac{ (0.39 \times1)^2}{0.1588}\\\\\alpha_c = \frac{ 0.1521}{0.1588}\\\\\alpha_c = 0.96 \;m/s^2[/tex]
e. To find the total acceleration, one second after the bug starts from rest:
Note: The total acceleration is the same as the resultant acceleration possessed by an object along a circular path.
[tex]T_a = \sqrt{\alpha_t^2 + \alpha_c^2 } \\\\T_a = \sqrt{0.39^2 + 0.96^2 }\\\\T_a = \sqrt{0.1521 + 0.9216 }\\\\T_a = \sqrt{1.0737 }\\\\T_a = 1.04 \; m/s^2[/tex]
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