rash219
- 22
- 0
Air Puck ! Correct Method or not...
An air puck of mass 0.25kg is tied to a string and allowed to revolve in a circle of radius 1.0m 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 1kg is tied to it. The suspended mass remains in equilibrium while the puck on the tabletop revolves
a. What is the tension in the string ?
b. What is the horizontal force acting on the string ?
c. What is the speed of the puck ?
\SigmaF = m * a
F_c = (m * v^2) / r
a_c = v^2 / r
a.
\SigmaF_y = m * a = 0
T - mg = 0
T = mg = 1.0kg * 9.8m/s^2 = 9.8N
b.
F_c = Tension on th string = 9.8N
c.
F_c = (m * v^2) / r = 9.8N
(0.25kg * v^2) / 1.0m = 9.8N
v = \sqrt{39.2} = 6.3 m/s
Is this correct...THANKS!
Homework Statement
An air puck of mass 0.25kg is tied to a string and allowed to revolve in a circle of radius 1.0m 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 1kg is tied to it. The suspended mass remains in equilibrium while the puck on the tabletop revolves
a. What is the tension in the string ?
b. What is the horizontal force acting on the string ?
c. What is the speed of the puck ?
Homework Equations
\SigmaF = m * a
F_c = (m * v^2) / r
a_c = v^2 / r
The Attempt at a Solution
a.
\SigmaF_y = m * a = 0
T - mg = 0
T = mg = 1.0kg * 9.8m/s^2 = 9.8N
b.
F_c = Tension on th string = 9.8N
c.
F_c = (m * v^2) / r = 9.8N
(0.25kg * v^2) / 1.0m = 9.8N
v = \sqrt{39.2} = 6.3 m/s
Is this correct...THANKS!
Last edited: