Does acceleration change in non-uniform circular motion?

AI Thread Summary
In non-uniform circular motion, the magnitude of acceleration consists of both tangential and centripetal components. The tangential acceleration remains constant while the centripetal acceleration changes as speed increases, following the formula a = v^2/r. The total acceleration is a combination of these two components. The centripetal force adjusts to accommodate changes in speed, but if the speed exceeds the force's capacity, sliding off can occur. Understanding these relationships is crucial for analyzing motion in circular paths.
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If an object is increasing speed at a constant rate around a circle, does the magnitude of acceleration change since there are both the tangential acceleration component (whose magnitude is constant) and the centripetal component (which changes in magnitude as the object increases in speed)?
 
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You could check whether speed appears in the formula for centripetal acceleration and in the formula for tangential acceleration, and consider how centripetal and tangential acceleration combine to make up the total acceleration.
 
The shape of the path depends on the strength of the centripetal force. a = v^2/r still holds for circular paths, even if there is a tangential acceleration. What happens is that the centripetal force accommodates the change. When driving a car and accelerating a curve, its because the static friction (centripetal force) is able to to increase to accommodate the increasing speed...although at some point it may not be able to handle it any further and you slide off.
 
Write the relation for resultant acceleration.
 
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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