A crankshaft with a diameter of 3.0 cm, rotating at 2200 rpm comes to a halt in 1.50 s . What is the tangential acceleration of a point on the surface of the crankshaft?

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A crankshaft with a diameter of 3.0 cm, rotating at 2200 rpm comes to a halt in 1.50 s . What is the tangential acceleration of a point on the surface of the crankshaft?

Physics
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2200rpm=2200 revolution per minutes \[\frac{ 2200 revolution }{ \min } \times \frac{ 1\min }{ 60seconds } \times \frac{ 2\pi radians }{ 1 revolution }\]=230.283 radians/seconds
\[\alpha=\frac{ v }{ t }\] Angular acceleration=velocity of circular motion divided by time. \[\alpha=\frac{ 230.283 radians/seconds }{ 1.50 seconds }\] =\[15.36 radians/seconds^2\]
\[a _{t}=\alpha r\] where tangential acceleration= angular acceleration x radius. Radius would have to be in meters. Radius=diameter/2

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The final answer was -2.30 m/s^2. I wouldn't have figured it out without your help, thank you so much.
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