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What is the linear acceleration of a point on the rim of a 36 cm diameter turntable which is turning at 33 rev/min?

My Work:

[tex]T = \frac{2 \pi r}{v}; v = \frac{2 \pi r}{T}[/tex]

[tex]a_{r} = \frac{v^{2}}{r}[/tex]

[tex]\frac{33 rev}{min} \times \frac{1 min}{60 sec} = \frac{.55 rev}{sec}[/tex]

[tex]\frac{2 \pi .36 m}{\frac{.55 rev}{sec}} = 4.1126 \frac{m}{sec}[/tex]

[tex]\frac{(4.1126\frac{m}{sec})^{2}}{.36 m} = 46.9826 \frac{m}{sec^{2}}[/tex]

Is my work correct? The final answer is suppose to be in [tex]\frac{m}{sec^{2}}[/tex]. The reason I ask is that the multiple choice answer that is nearest to my calculated answer is [tex]46.3 \frac{m}{sec^{2}}[/tex]. My professor admits that the exactness can be off a few times for the possible choices.

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# Homework Help: Linear Acceleration of a turntable

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