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anonymous

  • one year ago

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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  1. anonymous
    • one year ago
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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

  2. anonymous
    • one year ago
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    \[\alpha=\frac{ v }{ t }\] Angular acceleration=velocity of circular motion divided by time. \[\alpha=\frac{ 230.283 radians/seconds }{ 1.50 seconds }\] =\[15.36 radians/seconds^2\]

  3. anonymous
    • one year ago
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    \[a _{t}=\alpha r\] where tangential acceleration= angular acceleration x radius. Radius would have to be in meters. Radius=diameter/2

  4. anonymous
    • one year ago
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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.

  5. anonymous
    • one year ago
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    No problem!

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