Final answer:
The magnitude of the gravitational torque about the athlete's shoulder when holding a 2.5 kg ball with an outstretched 70 cm arm is 28.91 N·m. This torque is calculated by summing up the torques caused by the weight of the ball and the weight of the arm, considering the distance to the center of gravity for each.
Step-by-step explanation:
The student is asking about the gravitational torque around an athlete’s shoulder when holding a weight with an outstretched arm. To calculate the torque (τ), which is the rotational equivalent of force, we can use the following formula:
τ = r × F × sin(θ)
where r is the distance from the pivot point to the point where the force is applied (arm’s length), F is the force (in this case, the weight of the object, which is mass times the gravitational acceleration), and θ is the angle between the force and the lever arm (90 degrees or π/2 radians since the arm is held out to the side).
Let's first calculate the torque due to the ball:
For the 2.5 kg ball held at the end of a 70 cm arm (0.70 m), the force due to gravity on the ball is F = m × g (g = 9.8 m/s2), and the distance (r) is the full length of the arm:
Fball = 2.5 kg × 9.8 m/s2 = 24.5 N
τball = 0.70 m × 24.5 N × sin(90°) = 17.15 N·m
Now, for the arm's own weight, we need to consider the center of gravity of the arm, which is 30 cm (0.30 m) from the shoulder:
Farm = 4.0 kg × 9.8 m/s2 = 39.2 N
τarm = 0.30 m × 39.2 N × sin(90°) = 11.76 N·m
The total gravitational torque about the shoulder is the sum of the torques due to the ball and the arm:
τtotal = τball + τarm = 17.15 N·m + 11.76 N·m = 28.91 N·m
The magnitude of the gravitational torque about the shoulder when holding the arm straight out to the side, parallel to the floor, is 28.91 N·m.