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

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To find the tangential acceleration of a point on a crankshaft with a diameter of 3.0 cm that decelerates from 2500 rpm to a stop in 1.5 seconds, the relevant formula is at = Dv/Dt. The discussion highlights the need to convert angular velocity from rpm to radians per second and to apply rotational motion equations, specifically relating angular acceleration to tangential acceleration. Participants suggest using the initial and final angular velocities to calculate angular deceleration, which can then be used to determine tangential acceleration. Additionally, the number of revolutions made during the deceleration can be calculated using the angular displacement formula. The thread emphasizes the importance of understanding both linear and rotational motion equations for solving the problem.
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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
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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