An air puck of mass 0.674 kg is tied to a string and allowed to revolve in a cir

Question

An air puck of mass 0.674 kg is tied to a string and allowed to revolve in a circle of radius 1.05 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 1.10 kg is tied to it as seen in the figure.

a) The suspended mass remains in equilibrium while the puck on the tabletop revolves. What is the tension in the string?
1.08×101 N

b)What is the magnitude of the force which causes the centripetal acceleration of the puck?

c)What is the speed of the puck?

Explanation / Answer

a) For the equillibrium of the tied mass,

T- mg=0 ; T= 1.1*9.8=10.8 =1.08 x 10 N

b) Sine the string is not elastic, form newtons third law.

centripetal force(Fc)= T=10.8 N

c) Fc= mv2/r ; r-radius of the circular path and v-speed and mass(m)=0.674

v=4.1 ms-1

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