The unbalanced wheel O has a mass of 30 kg and is connected to a 20 kg counter weight block, B, with a continuous cable. If a constant moment of 15 N-m is a applied to the wheel in the clockwise direction, what is the force exerted on the wheel by the pin at center point, O? At the instance shown, the angular velocity, ω, is 4 rad/s. The center of mass, G is 5 cm from the center and the wheel's radius of gyration is 7 cm (about center point O). Assume the pulley is mass-less and frictionless.

There are two objects that are connected with the cable. Since the pulley is assumed mass-less, the tension in the cable does not vary.

The vertical equation of motion for the block is
     ΣF_y = ma_By
     T - 20(9.81) = 20a_By

Find the angular equation of motion for the wheel by summing all forces about O. Note, the wheel weight acts through the wheel's center of mass, and the moment of inertia, I, is related to the radius of gyration, k (I = mk²). Equating the moments to the angular acceleration gives,
     ΣM_O = I_O α = m(k_O)² α
     -15 - 30(9.81)(0.05) + 0.1T = 30(0.07)² α
     -29.72 + 0.1T = 0.1470α

There are three unknowns, T, α and a_By but only two equations. However, the angular velocity of the wheel and linear acceleration of the block are related,
     a_By = -r α = -0.1α

Combining all equations gives,
     -29.72 + 0.1 [20(-0.1α) + 20(9.81)] = 0.1470 α
     -10.10 = 0.3470 α
     α = -29.11 rad/s²
and
     T = 254.4 N

Now that all forces are known, the vertical and horizontal forces on the wheel can be summed,
     ΣF_x = ma_Gx = -mr_Gω²
     O_x - T = -30(0.05)4²
     O_x - 254.4 = -24
     O_x = 230.4 N

Vertical forces,
     ΣF_y = ma_Gy = mr_Gα
     O_y - 30(9.81) = 30(0.05)α
     O_y - 294.3 = 1.5 (-29.11)
     O_y = 250.6 N

Total reaction force at the center,
     O² = O_x² + O_y²

O = 340.4 N