Solving for Speed: Chip Dipped in Dip w/ Radius .5m

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SUMMARY

The discussion revolves around calculating the speed of a chip dipped into a dip with a circular radius of 0.5 meters, where a flavor crystal experiences twice the normal gravitational force. The initial calculation of 19.6 m/s, derived from multiplying the gravitational acceleration (9.8 m/s²) by 2, is incorrect as it confuses force with velocity. The correct approach involves using the centripetal force equation Fc=mv²/r and understanding that the net force acting on the chip is the difference between the forces due to gravity and the additional force applied.

PREREQUISITES
  • Understanding of centripetal force and its formula (Fc=mv²/r)
  • Basic knowledge of gravitational acceleration (9.8 m/s²)
  • Familiarity with kinematic equations for motion under constant acceleration
  • Concept of net force and its relation to mass and acceleration
NEXT STEPS
  • Study the derivation and application of the centripetal force equation (Fc=mv²/r)
  • Learn about kinematic equations for calculating velocity from acceleration and distance
  • Explore the concept of net force and its implications in dynamics
  • Investigate real-world applications of centripetal motion in fluid dynamics
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Students studying physics, particularly those focusing on mechanics and dynamics, as well as educators seeking to clarify concepts related to motion and forces.

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Homework Statement


a chip is dipped into dip and bottoms out in a circle with radius r. A flavor crystal on the chip is put under twice the normal gravity. If r = .5m, how fast is the chip traveling at the bottom of the dip. Neglect the mass of the chip.


Homework Equations





The Attempt at a Solution



My answer is taking 9.8 *2 which is 19.6 m/s. Is this correct? It seems to easy. But don't we need to take into account the mass of the chip?
 
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bigdaddy said:

Homework Statement


a chip is dipped into dip and bottoms out in a circle with radius r. A flavor crystal on the chip is put under twice the normal gravity. If r = .5m, how fast is the chip traveling at the bottom of the dip. Neglect the mass of the chip.


Homework Equations





The Attempt at a Solution



My answer is taking 9.8 *2 which is 19.6 m/s. Is this correct? It seems to easy. But don't we need to take into account the mass of the chip?
The acceleration due to gravity is 9.8 m/s2, not m/s and is not a velocity. In any case, "is put under twice the normal gravity" refers to a force not acceleration.

I frankly don't understand the problem- unless it is a very awkward way of asking what velocity something will have after starting from 0 velocity and moving 0.5 m at acceleration 19.6 m/s2. If that is the case, you need equations for velocity and distance with constant acceleration. How long will it take to go .5 m with constant 19.6 m/s2 acceleration? How will the velocity have changed in that time?

(Chip and dip? And the bowl of dip has a 50 cm radius? How big is the bottle of beer?)
 
I didnt understand the problem myself. I asked the guy who made the problem up (another student) and he explained what he was asking. now i understand. He meant to say basically it is a fluid motion from bag down through dip and up to mouth. the dipping motion has the .5m radius. He was saying Fc=mv^2/r ==> Fc=mg ==> netforce = 2mg-mg=mg. Only thing i don't understand now is what is the mass? Do i make something up? Another thing is that just tells me the centriputal force on the chip it doesn't really tell me how fast it is going.

Not just how big the beer is but what kind? also, i just figured out on this post how to get the ^ in. Dont think i have ever seen that on the keyboard. I know its 9.8m/s^2 i just didnt see that neat little symbol above the 6. haha
 

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