An object m of mass m=1kg is attached to a string that passes through a hole in the center of a table. The object m moves in a circle of radius r=1.3m. An object M of mass M=2.1kg is tied to the other end of the string, and hangs vertically beneath the table. Assuming the tabletop is frictionless, what speed must the object m have if the mass M is to remain at rest?
m g = M V^2/r
Solve for V. You know M, m, g and r.
To find the speed at which the object m must move so that the mass M remains at rest, we need to consider the forces acting on the system.
Let's start by analyzing the forces acting on the mass M. Since M is hanging vertically beneath the table, the only force acting on it is its weight, which is given by the equation F = mg, where g is the acceleration due to gravity (approximately 9.8 m/s^2). Therefore, the weight of M is equal to F = 2.1kg x 9.8 m/s^2 = 20.58 N.
Now, let's consider the forces acting on the mass m. As it moves in a circle, it experiences two primary forces: the tension in the string and the centripetal force. The tension in the string provides the centripetal force necessary to keep m moving in a circle. The tension in the string can be calculated using the equation T = mv^2 / r, where T is the tension, m is the mass of m, v is its speed, and r is the radius of the circle.
In this scenario, the tension in the string is equal to the weight of M (since M remains at rest). Therefore, we can equate the tension in the string to the weight of M: T = 20.58 N.
Substituting the known values into the equation T = mv^2 / r, we get:
20.58 N = (1 kg) x v^2 / 1.3 m.
Simplifying the equation, we rearrange it to solve for v:
v^2 = (20.58 N x 1.3 m) / 1 kg,
v^2 = 26.754 m^2/s^2.
Finally, we take the square root of both sides to find the speed:
v = √(26.754 m^2/s^2),
v ≈ 5.17 m/s.
Therefore, the object m must have a speed of approximately 5.17 m/s to keep the mass M at rest.