an air puck of mass m1= 0.25 kg is tied to a string and allowed to revolve in a circle of radius R=1.0m on a frictionless horizontal table . the other end of the string passes through a hole in the center of a table and a mass of m2=1.0kg is tied to it. the suspended mass remains in equilibrium while the puck on the tabeltop revolves.
a) what is the tension in the string
b) what is the horizontal force acting on the puck
c) what is the speed of the puck
the tension on the string has to be equal to the weight hanging down.
force on the puck equals tension.
force=m v^2/r solve for v
a) To find the tension in the string, we can consider the forces acting on the suspended mass (m2) in equilibrium.
The only force acting on m2 is the tension in the string, which provides the centripetal force required to keep m2 in circular motion.
The centripetal force (Fc) is given by the equation:
Fc = m2 * v^2 / R
Where:
m2 = mass of the suspended mass = 1.0 kg
v = velocity of the puck
R = radius of the circle = 1.0 m
Since m2 is in equilibrium, the tension in the string (T) is equal to the centripetal force:
T = Fc
b) The only horizontal force acting on the puck is the tension in the string. This force provides the centripetal force required to keep the puck in circular motion.
c) To find the speed of the puck, we can equate the centripetal force with the net force acting on the puck.
The net force (Fnet) is given by the equation:
Fnet = m1 * a
Where:
m1 = mass of the puck = 0.25 kg
a = acceleration of the puck
The centripetal force (Fc) is given by the equation:
Fc = m1 * v^2 / R
Since the net force is equal to the centripetal force:
m1 * a = m1 * v^2 / R
Rearranging the equation:
a = v^2 / R
Now, we can substitute the expression for acceleration in terms of velocity and radius back into the net force equation:
m1 * (v^2 / R) = m1 * v^2 / R
The mass cancels out, and we are left with:
v^2 = v^2 (no equation required)
This means that the speed of the puck will depend on the initial conditions or any other given information.
To solve these problems, we can use the concept of centripetal force. The centripetal force is the force that keeps an object moving in a circular path.
a) To find the tension in the string, we can equate the centripetal force to the tension in the string. The centripetal force is given by the equation:
F = (m1 * v^2) / R
Where m1 is the mass of the air puck, v is the speed of the puck, and R is the radius of the circle.
Since the puck is in equilibrium and not moving vertically, the tension in the string must also equal the weight of the suspended mass. This can be written as:
T = m2 * g
Where m2 is the mass of the suspended mass and g is the acceleration due to gravity.
Setting these two equations equal to each other, we get:
(m1 * v^2) / R = m2 * g
Solving for the tension T:
T = (m1 * v^2) / (m2 * g)
Substituting the given values of m1 = 0.25 kg, m2 = 1.0 kg, R = 1.0 m, and g = 9.8 m/s^2 into the equation will yield the tension in the string.
b) The horizontal force acting on the puck is the same as the tension in the string. So, the answer to part b) will be the same as the tension calculated in part a).
c) To find the speed of the puck, we can use the centripetal force equation mentioned in part a):
F = (m1 * v^2) / R
Rearranging the equation to solve for v:
v = sqrt((F * R) / m1)
Substituting the tension T calculated in part a), the radius R = 1.0 m, and the mass m1 = 0.25 kg into the equation will give the speed of the puck.
It is important to note that the answer to part a) and b) will be the same, as the tension in the string is the horizontal force acting on the puck. And the speed of the puck will depend on the calculated tension.