An air puck of mass m 1 = 0.32 kg is tied to a string and allowed to revolve in

Question

An air puck of mass m1 = 0.32 kg is tied to a string and allowed to revolve in a circle of radius R = 0.9 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 m2 = 1.0 kg is tied to it (see the figure below). The suspended mass remains in equilibrium while the puck on the tabletop revolves.

(a) What is the tension in the string?
N

(b) What is the horizontal force acting on the puck?
N

(c) What is the speed of the puck?
m/s

Explanation / Answer

Since the suspended mass remains in equilibrium, and the puck is revolving in a circle of constant radius, the puck must be moving at a constant speed. The net force which is causing the puck to move in a circular path at a constant speed is equal to the centripetal force.

Fc = 0.32 * v^2/0.9

The string connects the air puck to the object under the table. Since the table is frictionless, the tension in the string is the same at both ends of the string. At the end which is attached to the object under the table, the tension is supporting the weight of the object. This means the tension is equal to the weight of the object under the table.

a)

T = 1 * 9.81

= 9.81N

(b) The horizontal force on the puck (which is the centripetal force) is the same as the tension.

F(horizontal) = 9.8 N

(c) mv^2 / r = F(centripetal) ===> (0.32)(v^2) / (0.9) = 9.8

===> v = 5.25 m/s

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