You throw a Frisbee of mass m and radius r so that it is spinning about a horizontal axis perpendicular to the plane of the Frisbee. Ignoring air resistance, the torque exerted about its center of mass by gravity is

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Answer:

Torque τ =w ×0 = 0

Explanation:

We know that the torque is given by the product of the force and perpendicular distance between the force and the axis.

Here the gravity force act at the center and the rotational axis is also passing through the center.

Therefore the perpendicular distance between the force and the rotational axis would be zero.

Hence the torque will be

Torque = Force × Perpendicular distance

Torque = mg×0 = 0

Therefore the torque would be zero.

following are the response to the given question:

  • The torque is calculated as the product of the pressure and also the perpendicular distance between both the force and the axis.
  • In this case, gravity acts at the center, as well as the rotational axis likewise passes through it.
  • As a result, the perpendicular distance between both the force and the rotational axis is 0.

Calculating the torque:

[tex]\text{= Force} \times \text{Perpendicular distance}[/tex]

Therefore, the final torque would be '0'.

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