- #1

JrK

- 139

- 1

Hi :)

1/ First case

A wheel with a mass ##m## and a radius ##r## moves in horizontal translation and rotates around itself. The wheel is just above the ground, doesn't touch it. The wheel rotates CW if the wheel moves in translation to the right. The ground is horizontal and there is no gravity. The rotation of the wheel around itself is ##\omega## rd/s. The velocity of the translation of the wheel is ##v=\omega*r## m/s. The distance moved by the center of the wheel is exactly the distance moved by the rotation. From the ground, I apply a force ##F_1## on the wheel at the dot ##A##, the force ##F_1## is horizontal, parallel to the ground. The ground receives the force ##F_0##. For a small angle of rotation of the wheel, for example ##1°##, I don't need near any energy to move the force ##F_1## because the distance of translation of the wheel is the distance moved by the rotation of the wheel around itself. For example, for 1° with a radius of ##40## mm, the distance moved by the center of the wheel is ##0.7## mm and the distance moved by the force ##F_1## is ##0.000035## mm. With the values: ##F_1=F_4=F_0=F##. I win an energy of translation ##F*v*t## and I lost an energy from the rotation of the wheel ##F*r*\omega*t##, so the sum of energy is well 0. I drew for a difference of angle of 10° but I studied only for 1°:

2/ Second case

I use now the device I explained in the thread: https://www.physicsforums.com/threa...on-compared-to-the-energy-of-rotation.990326/. Like I studied the sum of energy in that device, we know the distance moved by the center of the wheel is exactly the distance moved by the rotation around itself, it is like the wheel of the first case 1/. The gears can cancel the torque from the force ##F_1## with the force ##F_2##, so there is no energy lost from the rotation: the gears prevent it. Like the sum of torques is 0, I can take a very low mass like I took in the device I studied before. The sum of forces on the center, on the horizontal axis (the vertical part of the sum of forces on the center cannot work because the wheel moves in horizontal translation), of the wheel is not ##0##, it is ##(1-\sqrt(2)/2)*F##, so the wheel wins an energy from the translation. The energy lost by the force ##F_1## in translation is very small compared to the energy wins by the center of the wheel, so I think the sum of forces on the center of the wheel is not like I think or the torque cannot be canceled by the gears. I drew the wheel at two positions, and a difference of 10°.

I apply the force ##F_1## on the dot ##A## to the wheel. The ground receives the force ##F_0##. At the dot ##B##, the gears cancel the torque from ##A##, the gears applies the force ##F_2## to the wheel. The rack receives the force ##F_5##. All the values of the forces from ##F_0## to ##F_5## is ##F##. The sum of forces on the center of the wheel is ##F_3+F_4##.

I can calculate the energy with equations, but for a small angle, I think it is not necessary because the difference of energy to move ##F_1## is very small compared to the energy wins by the center of the wheel in translation.

NB: In the device I studied with the gears before, I took the mass very small, with the wheel in the first case, I need a mass. But in the second case, like I cancel the torque from ##F_1## with the force from ##F_2##, I don't need the mass, and the mass can be very low to forget it.

1/ First case

A wheel with a mass ##m## and a radius ##r## moves in horizontal translation and rotates around itself. The wheel is just above the ground, doesn't touch it. The wheel rotates CW if the wheel moves in translation to the right. The ground is horizontal and there is no gravity. The rotation of the wheel around itself is ##\omega## rd/s. The velocity of the translation of the wheel is ##v=\omega*r## m/s. The distance moved by the center of the wheel is exactly the distance moved by the rotation. From the ground, I apply a force ##F_1## on the wheel at the dot ##A##, the force ##F_1## is horizontal, parallel to the ground. The ground receives the force ##F_0##. For a small angle of rotation of the wheel, for example ##1°##, I don't need near any energy to move the force ##F_1## because the distance of translation of the wheel is the distance moved by the rotation of the wheel around itself. For example, for 1° with a radius of ##40## mm, the distance moved by the center of the wheel is ##0.7## mm and the distance moved by the force ##F_1## is ##0.000035## mm. With the values: ##F_1=F_4=F_0=F##. I win an energy of translation ##F*v*t## and I lost an energy from the rotation of the wheel ##F*r*\omega*t##, so the sum of energy is well 0. I drew for a difference of angle of 10° but I studied only for 1°:

2/ Second case

I use now the device I explained in the thread: https://www.physicsforums.com/threa...on-compared-to-the-energy-of-rotation.990326/. Like I studied the sum of energy in that device, we know the distance moved by the center of the wheel is exactly the distance moved by the rotation around itself, it is like the wheel of the first case 1/. The gears can cancel the torque from the force ##F_1## with the force ##F_2##, so there is no energy lost from the rotation: the gears prevent it. Like the sum of torques is 0, I can take a very low mass like I took in the device I studied before. The sum of forces on the center, on the horizontal axis (the vertical part of the sum of forces on the center cannot work because the wheel moves in horizontal translation), of the wheel is not ##0##, it is ##(1-\sqrt(2)/2)*F##, so the wheel wins an energy from the translation. The energy lost by the force ##F_1## in translation is very small compared to the energy wins by the center of the wheel, so I think the sum of forces on the center of the wheel is not like I think or the torque cannot be canceled by the gears. I drew the wheel at two positions, and a difference of 10°.

I apply the force ##F_1## on the dot ##A## to the wheel. The ground receives the force ##F_0##. At the dot ##B##, the gears cancel the torque from ##A##, the gears applies the force ##F_2## to the wheel. The rack receives the force ##F_5##. All the values of the forces from ##F_0## to ##F_5## is ##F##. The sum of forces on the center of the wheel is ##F_3+F_4##.

I can calculate the energy with equations, but for a small angle, I think it is not necessary because the difference of energy to move ##F_1## is very small compared to the energy wins by the center of the wheel in translation.

NB: In the device I studied with the gears before, I took the mass very small, with the wheel in the first case, I need a mass. But in the second case, like I cancel the torque from ##F_1## with the force from ##F_2##, I don't need the mass, and the mass can be very low to forget it.

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