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
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
part a is the same as part b ;)
For B, the magnitude is simply the absolute value of what you got in part A.
To solve this problem, we can use Newton's laws of motion and principles of circular motion. Let's break down each part of the question to find the answers.
A) What is the tension T in the string?
To find the tension in the string, we need to consider the forces acting on the suspended mass m2. In equilibrium, the tension in the string is equal to the weight of the suspended mass. The weight is given by the formula:
Weight = mass * gravity
Where the gravity is the acceleration due to gravity (approximately 9.8 m/s^2).
So, the tension T is equal to m2 * g, where g is the acceleration due to gravity. Plugging in the values, we have:
T = 1.10 kg * 9.8 m/s^2
B) What is the magnitude of the horizontal force acting on the air puck?
To find the horizontal force acting on the air puck, we need to consider the forces acting on it. The only horizontal force acting on the puck is the tension in the string. This force provides the centripetal force necessary for circular motion. The centripetal force is given by the formula:
Centripetal force = mass * (velocity^2 / radius)
Since the puck is moving in a circular path, the centripetal force is equal to the horizontal force acting on the puck, which is the tension T. So, we have:
T = m1 * (velocity^2 / R)
Solving for the magnitude of the horizontal force, we get:
Horizontal force = m1 * (velocity^2 / R)
C) What is the speed of the puck?
To find the speed of the puck, we can use the equation for centripetal acceleration:
Centripetal acceleration = (velocity^2) / R
And since the puck is moving in a horizontal circle, this acceleration is related to the gravitational acceleration. We can set the centripetal acceleration equal to the gravitational acceleration:
(velocity^2) / R = g
Solving for velocity, we get:
velocity = sqrt(g * R)
Plugging in the values, we have:
velocity = sqrt(9.8 m/s^2 * 1.0 m)
By evaluating this expression, we can find the speed of the puck.