Question

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.

Answer 1

Answer:

Angular acceleration, α = 26.973 rad/s²

Explanation:

Given data:

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