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ryzh [129]
3 years ago
15

Suppose a yo-yo has a center shaft that has a 0.230 cm radius and that its string is being pulled.

Physics
1 answer:
Fofino [41]3 years ago
6 0

Answer:

Part a)

\alpha = 782.6 rad/s^2

Part B)

\omega = 587 rad/s

Part c)

a_t = 24.3 m/s^2

Explanation:

Part a)

As we know that

a = R \alpha

so we will have

a = 1.80 m/s^2

R = 0.230 cm

\alpha = \frac{a}{R}

\alpha = \frac{1.80}{0.230 \times 10^{-2}}

\alpha = 782.6 rad/s^2

Part B)

Angular speed of the yo-yo

\omega = \alpha t

so we have

\omega = 782.6 \times 0.750

\omega = 587 rad/s

Part c)

Tangential acceleration is given as

a_t = R \alpha

a_t = (3.10 \times 10^{-2})(782.6)

a_t = 24.3 m/s^2

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