A hockey puck of mass m = 70 g is attached to a string that passes through a hole in the center of a table, as shown in the figure below. The hockey puck moves in a circle of radius r = 1.20 m. Tied to the other end of the string, and hanging vertically beneath the table, is a mass M = 1.40 kg. Assuming the tabletop is perfectly smooth, what speed must the hockey puck have if the mass M is to remain at rest?
My physics optional extra work: This would be helpful so I can compare the solutions here as I do the work by myself.
Wouldn't it be better if you showed your work and let us check it for accuracy. That way we're checking you instead of you checking us.
To solve this problem, we can use the concepts of centripetal force and tension in the string.
First, let's analyze the forces acting on the system. We have two forces: the tension force in the string and the force of gravity acting on the hanging mass.
In order for the hanging mass M to remain at rest, the net force acting on it must be zero. Therefore, the tension force in the string must exactly balance the force of gravity.
The tension force in the string is responsible for providing the centripetal force required to keep the hockey puck moving in a circle. The centripetal force is given by the equation:
Fc = mv^2 / r
Where:
- Fc is the centripetal force
- m is the mass of the hockey puck
- v is the speed of the hockey puck
- r is the radius of the circle
In this case, the centripetal force is provided by the tension force in the string. So, we can equate the centripetal force to the tension force:
mv^2 / r = T
On the other hand, the force of gravity acting on the hanging mass is given by:
Fg = Mg
Where:
- Fg is the force of gravity
- M is the mass of the hanging mass
- g is the acceleration due to gravity
Since the hanging mass is at rest, Fg is equal to the tension force:
Mg = T
Now, we can substitute the value of T from the second equation into the first equation:
mv^2 / r = Mg
Solving this equation for v, we get:
v^2 = g * r * (M/m)
Finally, taking the square root of both sides:
v = √(g * r * (M/m))
Now we can substitute the given values:
g = 9.8 m/s^2
r = 1.20 m
M = 1.40 kg
m = 70 g = 0.070 kg
v = √(9.8 * 1.20 * (1.40 / 0.070))
Simplifying further:
v = √(9.8 * 1.20 * 20)
v ≈ 8.56 m/s
Therefore, the hockey puck must have a speed of approximately 8.56 m/s in order for the mass M to remain at rest.