An air puck of mass m1 = 0.45 kg is tied to a string and allowed to revolve in a circle of radius R = 1.0 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 m2 = 1.10 kg is attached to it. The suspended mass remains in equilibrium while the puck on the tables revolves.
A) What is the tension T in the string?
B) What is the magnitude of the horizontal force acting on the air puck?
C) What is the speed of the puck
4 answers
Teresa's class?
a)What is the tension T in the string?
in this case, tension = mass2 X gravity
so, tension = 1.10 kg X 9.8 m/s^2
tension = 10.78
For letter b, I am afraid I have no clue :(
c)What is the speed of the puck?
F = (m(v^2))/r
the tension is the force in this case, so:
tension = (m1(v^2))/r
v^2 = (tension X r)/m1
v^2 = (10.78 N X 1.0 m)/ .45 kg
v^2 = 23.95555556 m/s
Take the square root of both sides
v = 4.894441291 m/s
in this case, tension = mass2 X gravity
so, tension = 1.10 kg X 9.8 m/s^2
tension = 10.78
For letter b, I am afraid I have no clue :(
c)What is the speed of the puck?
F = (m(v^2))/r
the tension is the force in this case, so:
tension = (m1(v^2))/r
v^2 = (tension X r)/m1
v^2 = (10.78 N X 1.0 m)/ .45 kg
v^2 = 23.95555556 m/s
Take the square root of both sides
v = 4.894441291 m/s
part a is the same as part b ;)
For B, the magnitude is simply the absolute value of what you got in part A.
1 answer