Calculating Torque??

I don't really understand the scenario, but considering that marble can crack and has varying degrees of hardness, within one slab, from slab to slab, and depending on the region that it came from, I would not over torque. I would also find some kind of dampening washer/gasket so that you're not torquing metal directly to the marble..................maybe cork or rubber material between slab and fastener.

Neither do I, but he would need to specify the time needed to reach angular velocity of pi radians/second (even if he clarifies the other points)

Cheers.........Codey

It surprises many people but the torque needed to maintain an angular velocity has nothing directly to do with the moment of inertia. The key things that you need to clarify all relate to your coefficient of friction. Assuming that the static and kinetic coefficients are effectively the same, one now needs to know the normal force applied to the bearing to identify the torsional drag applied by the bearing. To find this out, one needs to know the relative relationship of the center of mass and the bearing's center of rotation. If they exactly coincide (a perfectly balanced wheel) then the only loading on the bearing that produces a torque drag is the pre-load and a fixed offset drag that has nothing to do with loading. (Normally this last component is so small that it is ignored.) When they don't precisely align one must know how much of a centripetal force is required to keep the offset mass in a uniform circular motion by the formula F=mv²/r with m being the mass of the slab, r the distance from the center of mass and axis of rotation, v the velocity of the mass (not the angular velocity). This now becomes the normal force that gets multiplied by the coefficient of friction that makes the opposing drag.

Now how quickly one wishes to accelerate to a particular angular velocity has everything to do with the moment of inertia.

actually i want to know how much power would be required to give pi radians per second angular velocity to the system. I have an idea that it would be done using the time required to reach that velocity and bearing friction. The slab is mounted on a shaft with perfect balance about its axis of rotation, so after reaching the required velocity, the system will consume torque to overcome bearing friction only.. i wanted to know how much torque would be required initially to start the system. let suppose time is 5 seconds :-P

Rotational equations of motion are isomorphic to linear equations of motion using Newton's laws. Isomorphisms:

mass m, kg <==> moment of inertia W, kg-m^2

velocity V, m/s <==> angular velocity w, radians/s (vector)

force F, Newtons or kg*m/s^2 <==> torque T, n-m or kg*m^2/s^2 (vector)

The question of what torque is required to give a given angular velocity is isomorphic to the question of what force is required to give a given velocity. Missing quantities in Newtonian mechanics are mass and time; moment of inertial and, isomorphically, mass, is given but we do have a missing degree of freedom.

We can use Newton's Second Law, F=(d/dt)[m*V] or, in rotational mechanics,

T=(d/dt)[W*w]

to get up to the missing variable of how much time is needed to achieve a given angular velocity. For the first-cut look at the problem, we usually assume constant acceleration, force, and torque over this time t; using this approximation gives us

T*t = W*w

or, the "average" torque is

T = (W*w)/t.

Frictional losses are found from the frictional torque, which is found from the weight on the shaft and the coefficient of friction mu of 0.2, and the radius of the shaft, rad=0.01 m. The frictional torque is

T.friction = r*m*g*mu

where g is gravitational acceleration and m*g is the weight on the shaft in newtons. Note that the problem statement does not include the weight on the shaft. The total torque would be the sum of these two torques.

you're right@Above

the weight of the shaft must also be included to find the frictional torque. also when the required angular velocity would be obtained, then the torque required would only be the frictional torque.

Thanks @good explanation

Cheers