Question

To develop muscle tone, a woman lifts a 2.50 kg weight held in her hand. She uses her biceps muscle to flex the lower arm through an angle of 60.0°.What is the angular acceleration (in rad/s2) if the weight is 24.0 cm from the elbow joint, her forearm has a moment of inertia of 0.250 kg · m2, and the net force she exerts is 550 N at an effective perpendicular lever arm of 2.10 cm? (Ignore gravity)How much work (in J) does she do?

Answer 1

To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:

$\tau = Fr$

Where,

F = Force

r = Radius

Replacing we have that,

$\tau = Fr$

$\tau = 21cm (\frac{1m}{100cm})* 550N$

$\tau = 11.55Nm$

The moment of inertia is given by 2.5kg of the weight in hand by the distance squared to the joint of the body of 24 cm, therefore

$I = 0.25Kg\cdot m^2 +(2.5kg)(0.24m)^2$

$I = 0.394kg\cdot m^2$

Finally, angular acceleration is a result of the expression of torque by inertia, therefore

$\tau = I\alpha \rightarrow \alpha = \frac{\tau}{I}$

$\alpha = \frac{11.55}{0.394}$

$\alpha = 29.3 rad/s^2$

PART B)

The work done is equivalent to the torque applied by the distance traveled by 60 °° in radians $(\pi / 3)$, therefore

$W = \tau \theta$

$W = 11.5* \frac{\pi}{3}$

$W = 12.09J$