A mass (M = 5 kg) is attached to a vertical axis by two light (i.e. massless) strings as shown. Each string has a length of 0.6 m, and the angle = 30° . If the system is rotated about the axis with frequency of 3 revolutions per second, what are the tensions in the two strings? (hint: they are not the same!)

Solution:

Choose a coordinate system with y=positive upwards and x=positive towards the center of the rotation, i.e. towards the axis. Draw a force diagram for the mass and apply F = Ma:

F_y = - Mg - T₂sin + T₁sin = Ma_y = 0

F_x = T₁cos + T₂cos = Ma_x =

Mv²/R

where we have used the fact that in circular motion a = v²/R, and that there is no acceleration in the y-direction. Also, we have

R = Lcos

where L = the length of each string. Now, we have two equations (the y and x equation above) in 2 unknowns (T₁ and T₂), assuming we know the speed v. The speed v is known, since we have the rate of revolution:

v = 2R/T

= 2Lcos/(1/3 s)

= 2(0.6 m)cos30°/(1/3 s)

= 9.79 m/s

We can solve the equations for the tensions by rearranging:

T₁ + T₂ = Mv²/Rcos

from the x-equation and

T₁ - T₂ = Mg/sin

from the y-equation. Adding these two equations together gives us T₁, and subtracting them gives us T₂:

T₁ = (M/2)[v²/Rcos + g/sin]

T₂ = (M/2)[v²/Rcos - g/sin]

Now plug in v=9.79 m/s, g=9.8 m/s², R=Lcos = (0.6m)cos30° = 0.52m, and M= 5 kg to these equations to get

T₁ = 582 N

T₂ = 484 N

Note that T₁ > T₂ since the upper string is the one counteracting gravity (through the vertical component of its tension).

Note: This was similar to problem 5-49 from the homework (the only difference being that we had two tensions instead of one normal force and one frictional force). It is essentially the same as problem 5-60 from the text (a level II problem).

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