Respuesta :
The complete question is;
To develop muscle tone, a woman lifts a 2.00-kg weight held in her hand. She uses her biceps muscle to flex the lower arm through an angle of 60.0º . What is the angular acceleration if the weight is 24.0 cm from the elbow joint, her forearm has a moment of inertia of 0.240 kg.m², and the net force she exerts is 759 N at an effective perpendicular lever arm of 2.00 cm?
Answer:
α = 42.76 rad/s²
Explanation:
We are given;
Mass; m = 2 kg
Distance; r = 24cm = 0.24m
Moment of inertia of the forearm; I = 0.24 kg.m²
Muscle Force; F = 759N
Perpendicular distance to lever arm; r' = 2cm = 0.02m
Angle at which she flexes arm; θ =60°
Let's assume that the arm starts extended vertically downwards. and we take the elbow joint as the point about which we calculate the torque.
Thus, we know that torque is given by the formula ;
τ = Force x Perpendicular distance
Thus, the toque exerted by force in the muscle is;
τ_muscle = 759 x 0.02 = 15.18 N.m
Also, torque exerted by the lifted weight is given as 0 because perpendicular distance is zero.
Thus, the net torque on the lower arm is;
τ_net = τ_muscle - τ_weight
τ_net = 15.18 - 0 = 15.18 N.m
Now, let's calculate moment of inertia of lifted weight.
The moment of inertia is given by;
I = mr²
Thus, moment of Inertia of lifted weight is; I_weight = 2 x 0.24² = 0.115 kg.m²
Thus,total moment of inertia is;
I_total = I_arm + I_weight
Thus, I_total = 0.24 + 0.115 = 0.355 kg.m²
Now, we can calculate the angular acceleration.
It's gotten from the equation;
τ_net = I_total•α
Where α is angular acceleration.
Thus making α the subject, we have ; α = τ_net/I_total
α = 15.18/0.355 = 42.76 rad/s²
