While a hammer thrower is rotating the hammer (weight and cable) in the circle, the instantaneous linear velocity of the hammer is 10 m/s. The radius of rotation for the hammer about the spinning axis is 1.5 m. If the mass of the hammer is 7.26 kg, what is the magnitude of the centripetal force that the athlete needs to produce to keep the hammer moving in this circular motion?

To solve this problem, we need to calculate the centripetal force required to keep the hammer moving in a circular path. The centripetal force is the force that keeps an object moving in a circle, and it is directed towards the center of the circular path. The formula for centripetal force [tex]F_c[/tex] is given by:

[tex]F_c = \frac{m \, v^2}{r}[/tex]

where:

[tex]m[/tex] is the mass of the object,
[tex]v[/tex] is the instantaneous linear velocity,
[tex]r[/tex] is the radius of the circular path.

Given:

Mass of the hammer, [tex]m = 7.26[/tex] kg,
Instantaneous linear velocity, [tex]v = 10[/tex] m/s,
Radius of rotation, [tex]r = 1.5[/tex] m.

Now, substituting these values into the formula:

[tex]F_c = \frac{7.26 \, \text{kg} \times (10 \, \text{m/s})^2}{1.5 \, \text{m}}[/tex]

[tex]F_c = \frac{7.26 \, \text{kg} \times 100 \, \text{m}^2/\text{s}^2}{1.5 \, \text{m}}[/tex]

[tex]F_c = \frac{726 \, \text{kg} \cdot \text{m/s}^2}{1.5}[/tex]

[tex]F_c = 484 \, \text{N}[/tex]

Therefore, the magnitude of the centripetal force that the athlete needs to produce to keep the hammer moving in its circular motion is 484 Newtons.