Physics forces problem

Question

A block of mass m1 is attached to a cord of length L1, which is fixed at one end (in the center of the circular trajectory). The block moves in a horizontal circle on a frictionless table. A second block of mass m2 is attached to the first by a cord of length L2 and also moves in a circle (on the same frictionless table), as shown in the figure below. Each block moves with a constant speed. The two cords remain parallel and the blocks have the same period of motion T.

Algebraically solve for the tensions T1 and T2 in each cord. Determine the values of the tensions for the following set of parameters: m1 = 1.0 kg, m2 = 2.0 kg, L1 = 1.0 m, L2 = 2.0 m, and T = 2.0 s.

Answer 1

This is a great question Maryn. I have been brushing up on physics during this time period.

I hope there is a way to make and, or draw diagrams here!!!!!

You have to diagram everything to hold and see the concepts.

So, draw yourself a diagram. Then we know automatically that this involves F_cp. And, you have stated very precisely and concisely all the other parameters.

Our eqns:

F_cp=m(v²/r)

But, we're missing the linear velocity. But, no worries, from trig we have angular velocity: ω_avg= θ/t, linear velocity: v = rω

Our period, T = 2s, thus, ω_avg= θ/t = 2π/2s = π rad/s

Now from angular v to linear v (we have to do this twice):

For our first object, let's call it m₁: it has an r = 1.0 m v₁ = rω = 1.0m(π) =π m/s

Now, the 2nd object, let's call it m₂: make sure to add the L₁ and L₂!!! therefore, r = 2 + 1 = 3.0 m

v₂ = rω = 3m(π) = 3π m/s

Now we can use our F_cp eqn. *Trick* for the 1st object you have to add both masses in order to determine T₁!!! And, the 2nd object add the L (or r)

Therefore, T₁ = (m₁ + m₂)(v₁²/r) = (1 + 2)(π²/1) = 3π² N

T₂ = m₂(v₂²/r) = 2((3π)²/3) = 6π² N

Double check my work, I'm distracted by a couple things and it's been a while since I've done physics