Newton's Third Law/ Circular Motion

[dekoi]

Question:
Mass m1 on the frictionless table is connected by a string through a hole in the table to a hanging mass m2. With what speed must m1 rotate in a circle of radius r if m2 is to remain hanging at rest?

My answer:
Fnetym2 = T - m2g
For m2, a = 0
Therefore,
0 = T - m2g
T = m2g

Fnetxm1 = T
m1a = T
m1a = m2g
Therefore,
a = m2g / m1

Since a = v^2 / r, v = sqrt(ra)

Therefore,
v = sqrt(rm2g / m1)

Is this correct?

 

[Päällikkö]

Looks correct to me.

 

[quicknote]

Yes, your solution is correct. This is an application of Newton's Third Law, which states that for every action, there is an equal and opposite reaction. In this case, the tension force in the string pulling on m1 is equal to the weight of m2 hanging on the other side. This results in a circular motion for m1 with a centripetal acceleration equal to the weight of m2 divided by the mass of m1. By equating this with the equation for centripetal acceleration, you have correctly solved for the required speed. Great job!