Determine the tension in the cord:
Two blocks containing the weights and positions illustrated in Figure rest on a frame which rotates around its vertical axis at a constant speed. The coefficient of friction among the blocks and the frame is 0.20 neglecting the weight and the friction of the pulley. Compute the speed in rpm at which the blocks shall start to slide. Also determine the tension in the cord at that instant.
Solution
As the system starts rotating around the vertical axis, acceleration equal to r ω2 will be acting on each block. The outer block shall have magnitude of acceleration more and has a mass larger than the other block. Thus, we will assume that outer block will have a motion oriented outside as illustrated on FBD. This will cause the inner block to move towards axis of rotation.
We have N1 = 5 × 9.81 = 49.05 N
And F = μ N1 = 0.2 × 49.05
= 9.810 N
By considering equilibrium ∑ Fx = 0, T + F = 5 × an
an = r ω2 = 1.8 ω2
∴ T = 5 × 1.8 ω2 - 9.81
= 9 ω2 - 9.81 ----------- (1)
Let FBD of other block
F = 0.2 × 3 × 9.81 = 5.89 N
∑ Fx = 0 ∴ T = Fi + F
= 3 × 1.2 ω2 + 5.89
= 3.6 ω2 + 5.89
∴ We have, T = 9 ω2 - 9.81 = 3.6 ω2 + 5.89
9 ω2 - 3.6 ω2 = 5.89 + 9.81 ----------- (2)
∴ ω2 =( 9.81 + 5.89 )/5.4 = 2.907
ω = 1.705 rad / sec
∴ N = (ω× 60) /2 π = 16.3 rpm
Substituting this value of ω in one of the equations for T, we obtain
T = 9 ω2 - 9.81 = 16.35 N