# Force acting during rotating a disc

• ussername
In summary: W}{dt}=(\boldsymbol F,\boldsymbol v_A)+(\boldsymbol\omega,\boldsymbol M_A)$$Now let's assume that the body (you) has a uniform mass distributed uniformly along the rotation axis. In this case, the torque and angular momentum of the disk are zero vectors. ussername Take a disc that can rotate with respect to the rotation axis. For simplicity, let's assume that its mass is homogeneously distributed along the rotation axis, and gravity and frictional forces do not act on the disk. In the first case, the disc does not rotate. All elements of the disk have zero force elements. The torque and angular momentum of a disc are zero vectors. In the latter case, a man rotates the disc with his arm. This means that a person acts on the disc elements with non-zero force elements that are perpendicular to the elements positioning vectors from the rotation axis. Thus, when the disc is rotated, there is a nonzero torque of the disc, that is, the angular momentum of the disk is changing. After spinning the disc with nonzero angular velocity, only elements of centrifugal forces that are parallel to the position vectors act on all of the disc elements, and the resultant torque of disc is zero. The angular momentum of the rotated disc is not changing, and it is a non-zero vector. Are these considerations correct? When I imagine that a person acts on the disk with a tangential force, I find that the elements of the tangential forces of every two points of the disc with the opposite position vectors are substracted, and the total force acting on the disk is zero. Is that so? Why? Some of your observations might be correct and some not but your description of the problem is so rambling and unscientific that it is hard to tell for sure . Please sort out your ideas and then post a nice concise description of the problem with clear diagrams so that we can discuss this matter properly . The physics of spinning disks is quite an interesting topic to explore . Ok let's discuss the force distribution when rotating the disc with your arm: ussername said: When I imagine that a person acts on the disk with a tangential force, I find that the elements of the tangential forces of every two points of the disc with the opposite position vectors are substracted, and the total force acting on the disk is zero. Is that so? But from my experience arm muscles should do nonzero mechanical work dW=F*dl when rotating disc so the overall force could not be zero from this point of view. Let ##A## be a body fixed point. Then$$\frac{dW}{dt}=(\boldsymbol F,\boldsymbol v_A)+(\boldsymbol\omega,\boldsymbol M_A);
here ##\boldsymbol F## is a net force applied to the rigid body;
##\boldsymbol v_A## is velocity of the point A;
##\boldsymbol\omega## is angular velocity;
##\boldsymbol M_A## is net torque about the point A.

Now for the point A choose the center of your disk

## 1. What is the force that acts during the rotation of a disc?

The force that acts during the rotation of a disc is known as centripetal force. This is the force that keeps an object moving in a circular path and is directed towards the center of the circle.

## 2. How is the centripetal force calculated?

The centripetal force can be calculated using the formula F = mv^2/r, where m is the mass of the object, v is the velocity, and r is the radius of the circle.

## 3. What happens if the centripetal force is not balanced?

If the centripetal force is not balanced, the object will not be able to maintain its circular motion and will either move towards the center or fly off in a straight line.

## 4. Can the centripetal force be increased?

Yes, the centripetal force can be increased by either increasing the mass of the object, increasing the velocity, or decreasing the radius of the circle.

## 5. What are some real-life examples of centripetal force?

Some examples of centripetal force in everyday life include the rotation of a merry-go-round, the motion of a satellite around the Earth, and the spin of a washing machine during the rinse cycle.

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