In which of the following cases is the torque about the shoulder due to the weight of the arm the greatest? case 1: a person holds her arm at an angle of 30 above the horizontal (hand is higher than shoulder). case 2: a person holds her amr straight out parallel to the ground. case 3: a person holds her arm at an angle of 30o below the horizontal (hand is lower than shoulder).

if the person lets her arm swing freely from an initial straight-out parallel-to-the-ground position, when is the angular acceleration of the arm about the shoulder the greatest?
case 1: immediately after her arm begins to swing.
case 2: when the arm is vertical.
case 3: the angular acceleration is constant.

a) Case 2 b) Case 1

Explanation:

a) By definition, the magnitude of a torque, referred to a given point, is expressed as the product of the force that causes the torque, times the perpendicular distance to the reference point.

If we assume that the only force acting on the arm is the weight of the arm, and that this is concentrated in a point in the center of it (taking the arm as a solid bar with the center of mass at the mid-point), clearly the torque will be the greatest when the force be exactly perpendicular, which is the case of the arm placed straight out parallel to the ground (Case 2).

b)  As the torque and the angular acceleration are directly proportional each other (being the rotational inertia the proportionality constant) the angular acceleration will be maximum when torque be maximum also, which is the case that the arm begins to swim, due to the perpendicular distance to the shoulder is the maximum possible (Case 1).

1/0.14nm = 10^10/1.4 a lot