An air puck of mass 0.19 kg is tied to a string and allowed to revolve in a circle of radius 0.85 m on a frictionless horizontal table. The other end of the string passes through a hole in the center of the table, and a mass of 0.437 kg is tied to it as shown above. The suspended mass remains in equilibrium while the puck on the tabletop revolves. What is the speed of the puck?

asked by Anonymous

8 years ago

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1 answer

mg = ma =mv^2/r
.437*9.8 = 0.19*v^2/.85

answered by Chanz

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Question

An air puck of mass 0.19 kg is tied to a string and allowed to revolve in a circle of radius 0.85 m on a frictionless horizontal table. The other end of the string passes through a hole in the center of the table, and a mass of 0.437 kg is tied to it as shown above. The suspended mass remains in equilibrium while the puck on the tabletop revolves. What is the speed of the puck?

1 answer

mg = ma =mv^2/r
.437*9.8 = 0.19*v^2/.85

Chanz · Answer

mg = ma =mv^2/r .437*9.8 = 0.19*v^2/.85