String in vertical circular motion

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In vertical circular motion, an object tied to a string does experience changes in speed due to the influence of gravity. While the average speed may be constant, the actual speed varies at different points in the motion, particularly at the top and bottom of the circle. This phenomenon is evident in scenarios like roller coasters, where speed decreases on uphill segments and increases on downhill segments. The concept of constant speed applies primarily to horizontal circular motion, where only direction changes without speed variation. Therefore, speed changes in vertical circular motion are a critical aspect to understand.
BassMaster
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If an object is tied to a string in vertical circular motion (yes, I mean vertical as in the string has the force of gravity acting on it), does it actually change speed? I was given a question where the 'average speed' of an object in vertical circular motion was 6.1 m/s, and then they wanted me to find the point of least speed. So I guess speed changes then? But why? I was taught that the speed of an object in circular motion is always the same but the object is accelerating because direction changes. Does this ONLY apply to horizontal circular motion?
 
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no sir, yes sir. The speed in say a loop de loop roller coaster most certainly changes with uphill or downhill part of the trajectory. Yet its still executing circular motion.
 
Alright, thanks for your help.
 
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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