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Suppose a yo-yo has a center shaft that has a 2.70 cm radius and that its string is being pulled. If the string is stationary and the yo-yo accelerates away from it at a rate of 1.65 m/s2, what is the angular acceleration of the yo-yo

User Flamey
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1 Answer

2 votes

Answer:


\alpha= (1.65 m/s^2)/(0.027 m)= 61.11 rad/s^2

Step-by-step explanation:

For this case we can use the following formula for the angular velocity:


w = w_o + \alpha t

where
w represent the final angular velocity ,
w_o the initial angular velocity , t the time and
\alpha the angular acceleration.

And for the linear acceleration we have this formula:


a = r \alpha

Where a represent the linear acceleration and
\alpha the angular acceleration.

For this case the linear acceeleration is given
a_c = 1.65 m/s^2

And the radius of the yo-yo is also given
r= 2.7 cm= 0.027 m

So then we can use the following formula:


a = r \alpha

If we replace the values we got:


1.65m/s^2 = 0.027 m (\alpha)

And solving for
\alpha we got:


\alpha= (1.65 m/s^2)/(0.027 m)= 61.11 rad/s^2

User Karrin
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