Normal forces for small car performing a vertical loop

AI Thread Summary
The discussion revolves around calculating the normal force exerted on a small car traveling in a vertical loop. The car has a mass of 0.800 kg and is said to travel at a constant speed of 12 m/s, but this speed is later deemed incorrect. The normal force at the top of the loop is given as 6.00 N, and the centripetal acceleration calculated is 17.31 m/s². The participants are focused on determining the normal force at the bottom of the track, emphasizing the need for clarity in the calculations. The conversation highlights the importance of accurate values in physics problems to avoid confusion.
Anatalbo
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Homework Statement



A small car with mass .800 kg travels at a constant speed of 12m/s on the inside of a track that is a vertical circle with radius 5.0m. If the normal force exerted by the track on the car when it is at the top of the track is 6.00N, what is the normal force at the bottom of the track?

Reference https://www.physicsforums.com/threa...mics-car-traveling-in-vertical-circle.656316/

Homework Equations

The Attempt at a Solution


I got that the acceleration towards the center of the circle is 17.31 m/s^2
 
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Greg Bernhardt said:
Please show your work.
N+w=FC (normal force, weight, centripetal force)

6N+(9.81)(0.8)=0.8a
a=17.31 m/s^2
 
Anatalbo said:
N+w=FC (normal force, weight, centripetal force)

6N+(9.81)(0.8)=0.8a
a=17.31 m/s^2
Ignore the given constant speed of 12m/s, as it is incorrectly stated as such, which you will discover by reading the thread you referenced. So you now have the correct centripetal acceleration. So now show how you would calculate the normal force at the bottom, assuming the car's speed is constant.
 
PhanthomJay said:
Ignore the given constant speed of 12m/s, as it is incorrectly stated as such, which you will discover by reading the thread you referenced. So you now have the correct centripetal acceleration. So now show how you would calculate the normal force at the bottom, assuming the car's speed is constant.
Thanks! I was confused about why the values didn't add up
 
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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