Rotation Problem: Angular Displacement & Spin Time Calculation

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The discussion revolves around calculating the angular displacement and time until a spinning coin stops. The coin initially spins at 5 rotations per second, equivalent to 10π radians per second, and experiences a deceleration of 0.4 r/s². The problem involves determining the angular displacement when the coin's spin rate decreases to half its original velocity, with the expected answer being 924 radians. Participants clarify that while the coin slows down, the angular velocity does not become negative; instead, the angular acceleration is negative. The calculations and equations used in the problem highlight the importance of correctly applying physics principles to avoid errors in results.
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Homework Statement



A coin is spinning on its edge at 5 rotations per second.
Friction slows down its spin rate at .4 r/s2

a) what angular displacement does the coin have by the time it's slowed down to half its original angular velocity.

b) how long before the coin stops spinning?
 
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5 rev per second = 10 pi radians per second

\alpha = 0.4 r/s2
ωi (angular velocity initial) = 10 pi r/s
ωf (angular velocity final) = 5 pi r/s

Δθ = angular displacement

Known equations

ωf2 = ωi2 + 2\alpha(Δθ)

I tried plugging the know variables in but i keep getting a negative answer.

btw answer is supposed to be 924 Rad.
 
KTiaam said:
5 rev per second = 10 pi radians per second

\alpha = 0.4 r/s2
Is it getting faster or slower?
 
haruspex said:
Is it getting faster or slower?

slower. that does not make velocity negative though?
 
It'll make alpha negative.
 
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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