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?
mg = ma =mv^2/r
.437*9.8 = 0.19*v^2/.85
To calculate the speed of the puck, we can use the concept of centripetal force.
The centripetal force acting on an object moving in a circular path is given by the equation:
Fc = (mv^2) / r
Where:
Fc is the centripetal force
m is the mass of the object
v is the velocity (or speed) of the object
r is the radius of the circular path
In this case, the centripetal force is provided by the tension in the string, as there is no friction or any other external force acting on the puck.
We can equate the centripetal force with the tension in the string:
Tension = Fc
Now, we need to find the tension in the string. To do that, we can use the equilibrium condition. Since the suspended mass is in equilibrium, the tension in the string equals the weight of the suspended mass (mg):
Tension = mg
We can calculate the tension using the mass of the suspended mass (0.437 kg) and the acceleration due to gravity (9.8 m/s^2):
Tension = (0.437 kg) * (9.8 m/s^2)
Now, we can equate the tension to the centripetal force:
Tension = Fc
(0.437 kg) * (9.8 m/s^2) = (0.19 kg) * (v^2) / (0.85 m)
Simplifying the equation:
(0.437 kg) * (9.8 m/s^2) * (0.85 m) = (0.19 kg) * (v^2)
Now, we can solve for the velocity (v):
v^2 = [(0.437 kg) * (9.8 m/s^2) * (0.85 m)] / (0.19 kg)
Taking the square root of both sides to solve for v:
v = sqrt{[(0.437 kg) * (9.8 m/s^2) * (0.85 m)] / (0.19 kg)}
Evaluating the expression gives the speed of the puck.