AP Physics- Cup and Tray Problem A cup of water is swung in a circle on a servin

Question

AP Physics- Cup and Tray Problem A cup of water is swung in a circle on a serving tray; the radius of the circle is 1.00 m. What is the minimum speed that the tray/cup would have to move at the top of the vertical circle in order for the cup not to fall off of the tray? (Also, why does the water not leave the cup and why does the cup not leave the tray?) What is the "apparent weight" of the cup of water when the tray is at the top of the vertical circle? If the tray/cup continue to travel in a vertical circle at the same speed calculated in part b. What is the "apparent weight" of the cup of water when the tray is at the bottom of the vertical circle? The tray is now spun in a horizontal circle, that is the tray is vertical... If the coefficient of static friction between the cup and the tray is 0.300, what is the minimum speed with which the cup must move? What is the "apparent weight of the cup?

Explanation / Answer

a) let minimum velocity be v

centripetal force at top = mv^2/r = mg

   v= sqrt(rg)

    = sqrt(1*9.8)

   = 3.13 m/s

It does not fall because only vertical force mg acts as centripetal force.

b) apparent weight = 0 N

c) Apparent weight = 2mg

d) limiting speed = v

    mv^2/r = umg

    v = sqrt(urg)

    = sqrt(0.3*1*9.8)

    = 1.715 m/s

e) apparent weight = mg*sqrt(1^2+0.3^2) = 1.044 mg

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