Suppose a yo-yo has a center shaft that has a 0.230 cm radius and that its string is being pulled.
(a) If the string is stationary and the yo-yo accelerates away from it at a rate of 1.80 m/s2, what is the angular acceleration of the yo-yo in rad/s2? rad/s2.
(b) What is the angular velocity in rad/s after 0.750 s if it starts from rest? rad/s.
(c) The outside radius of the yo-yo is 3.10 cm.What is the tangential acceleration in m/s2 of a point on its edge? m/s2

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aristocles aristocles · Answer 1 · 2019-11-25T02:30:02+01:00

Answer:

Part a)

[tex]\alpha = 782.6 rad/s^2[/tex]

Part B)

[tex]\omega = 587 rad/s[/tex]

Part c)

[tex]a_t = 24.3 m/s^2[/tex]

Explanation:

Part a)

As we know that

[tex]a = R \alpha[/tex]

so we will have

[tex]a = 1.80 m/s^2[/tex]

[tex]R = 0.230 cm[/tex]

[tex]\alpha = \frac{a}{R}[/tex]

[tex]\alpha = \frac{1.80}{0.230 \times 10^{-2}}[/tex]

[tex]\alpha = 782.6 rad/s^2[/tex]

Part B)

Angular speed of the yo-yo

[tex]\omega = \alpha t[/tex]

so we have

[tex]\omega = 782.6 \times 0.750[/tex]

[tex]\omega = 587 rad/s[/tex]

Part c)

Tangential acceleration is given as

[tex]a_t = R \alpha[/tex]

[tex]a_t = (3.10 \times 10^{-2})(782.6)[/tex]

[tex]a_t = 24.3 m/s^2[/tex]