Find the angular acceleration produced given the mass lifted is 12 kg at a distance of 27.1 cm from the knee joint, the moment of inertia of the lower leg is 0.959 kg m2 kg m 2, the muscle force is 1504 1504 N, and its effective perpendicular lever arm is 3.3 3.3 cm.

Angular acceleration, α = 26.973 rad/s²

Explanation:

Given dа ta:

Lifted mass, M = 12 kg

Distance of the lifted mass = 27.1 cm = 0.271 m

Effective lever arm, d = 3.3 cm = 0.033 m

Moment of inertia, I = 0.959 kg. m²

Applied force, F = 1504 N

Now,

the torque (T) is given as:

T = F * d

also,

T = I * α

where,

α is the angular acceleration

Now,

Total moment of inertia, I = 0.959 + 12 * (0.271) ² = 1.840 kg. m²

Now equation both the torque formula and substituting the respective values, we get

1504 * 0.033 = 1.840 * α

⇒ α = 26.973 rad/s²