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Assuming i have a circular disc of a significant mass and diameter. Neglecting friction, if a motor is put at its center how do i calculate the torque required for the wheel to just start to rotate?

Since the center of gravity is at its middle point, the only vectors i can work with dont provide any moment

i've seen in posts such as this saying a very low torque can rotate the wheel but how low?

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    $\begingroup$ Assuming no friction, any torque will start the disc spinning. It might accelerate very slowly however. $\endgroup$
    – Eric S
    Feb 3 at 3:56
  • $\begingroup$ lets consider a pulley drive system with a load force applied tangential to the disc. The torque required will be that force times the radial distance. If load force was non existent what will be the new component of force that will need to be overcome based on the properties of the disc? If for example the center of gravity was not at the midpoint then the torque to start spinning the disc will be = mg × distance from rotational axis. right? so we can safely assume that no other force component acts outside the center of gravity for the disc? $\endgroup$
    – Tony
    Feb 3 at 8:34
  • $\begingroup$ I think you're missing some understanding on "moment of inertia". This should eliminate your confusion with "centre of gravity" and addresses solid wheels, spoked wheels and all other variations of asymmetrical shapes.. $\endgroup$
    – Transistor
    Feb 3 at 9:45
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let me take a stab at this.

Similar to linear motion of a mass m on a frictionless surface or in space, that a force $F$ no matter how small will move a mass no matter how big, albeit with smaller acceleration if the mass is bigger or F smaller.

$$F=m\alpha$$

A torque no matter how small will turn a disk or a randomly shaped piece of rock or an ice skater or even a massive cloud of plasma as those that turn around usually a black-hole center in space and later become galaxies. you toss a hammer in the air and the slightest pitch of your wrist will force it to spin around an unintuitive center of rotation.

The equivalent of mass for rotation is second area moment or moment of inertia of the object which is related to how far its mass is distibuted from its CG, and equivalent of linear acceleration is angular acceleration, and the force changes to torque.

$$\tau= I \alpha$$

For example, in order to measure G, the famous newton's gravity factor they suspended a wide, heavy dumbbell from a long string and after it was carefully set to stand still, it started to turn when they placed a massive object near one end of it and thus they measured G. And we know G is really a very small number.

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In your case it is the inertia that you have to calculate with. A quantity related to inertia is rotational inertia (→ moment of inertia) See Wikipedia for more info.https://en.wikipedia.org/wiki/Inertia

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  • $\begingroup$ Thanks for the reply. Inertia is resistance to change in motion but in this case there is no motion of the body yet. So the torque i would be calculating with rotational inertia would be torque required for changing momentum instead of torque to start rotation $\endgroup$
    – Tony
    Feb 3 at 8:53
  • $\begingroup$ @Tony: "So the torque i would be calculating with rotational inertia would be torque required for changing momentum instead of torque to start rotation." They're the same except that in the case of starting the initial momentum is zero. $\endgroup$
    – Transistor
    Feb 3 at 9:41

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