What is the tangential acceleration of a point on a rotating crankshaft?

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SUMMARY

The discussion centers on calculating the tangential acceleration of a point on a crankshaft with a diameter of 3.0 cm, initially rotating at 2500 rpm and coming to a stop in 1.5 seconds. The relevant formula for tangential acceleration is at = Dv/Dt. To find the angular acceleration (thetadoubledot), users should utilize the equations for rotational motion, specifically the relationship between angular velocity (thetadot) and time.

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Homework Statement


A 3.0-cm diameter crankshaft that is initially rotating at 2500 rpm comes to a halt in 1.5 s. What is the tangential acceleration of a point on the surface of the crankshaft? How many revolutions does the crankshaft make as it comes to a stop?


Homework Equations



at = Dv/Dt


The Attempt at a Solution



honestly I am really stuck on this because the only formula that I can find is this one. Any ideas on where to head with it?
 
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For linear motion we have,

v = v_o + a*t

For rotational motion we have,

thetadot = thetadot_o + thetadoubledot*t

thetadot = d theta/dt thetadoubledot = d^2 theta/dt^2

You know thetadot at t = o and you know thetadot at t = 1.5 seconds, solve for thetadoubledot
 

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