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) 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).

ConcepcionPetillo ConcepcionPetillo · Answer 2 · 2022-03-11T13:33:36+01:00

The torque is maximum in Case 2 and the angular acceleration is maximum in Case 1: Immediately after her arm begins to swing.

Torque:

(a) The torque is the vector product between the force acting on the body and its position vector from the reference point.

mathematically, the torque is :

τ = r × F

τ = rFsinθ

where θ is the angle between the position vector r and the force F

If we assume the stretch of the arm as a position vector, then the gravitation force F = mg on the arm will be acting downward.

Case 1 : when the arm is 30° above horizontal:

τ = rFsin(90+30)

τ = 0.86rF

Case 2: when the arm is level with horizontal:

τ = rFsin(90)

τ = rF

Case 3: when the arm is 30° below the horizontal:

τ = rFsin(90-30)

τ = 0.86rF

So the torque is maximum in Case 2.

(b) As the torque and the angular acceleration are directly proportional to each other:

[tex]\tau = I\alpha[/tex]

the angular acceleration (α) will be maximum when torque be maximum also, which is the case that the arm begins to swing, due to the perpendicular distance to the shoulder being the maximum possible. So it will be maximum immediately after her arm begins to swing.

Learn more about torque:

https://brainly.com/question/13795320?referrer=searchResults

https://brainly.com/question/3388202?referrer=searchResults