Check out some similar questions!

hello94857 · Feb 2, 2014, 06:36 AM

If there is an symmetrical 8kg weight with a radius of 25cm hanging from a piece of thin string with a 75cm lever protruding from the side at a 90 degree angle therefore horizontal,

How much force need be applied to the end of the 75cm lever to maintain a 2000rpm momentum for the 8kg weight?

ebaines · Feb 2, 2014, 07:25 AM

Not enough information is provided to give an answer. Once the object is spinning it takes little to no force tp keep it spinning (ignoring air friction and friction of the string). Applying force to the lever arm imparts a torque on the weight, which causes it to speed up. So if the weight is initially stationary the force will cause it to start rotating. But Newton's 1st law ( a body in motion tends to stay in motion) says once spinning at 2K RPM no additional force is needed. So to answer your question you need to specify:

1) How quickly (in seconds) you want the weight to spin up from 0 RPM to 2000 RPM.
2) Can you quantify the opposing forces of friction in the mechanism?
3) You say the weight is symmetric - but is it constant across its width (like a flat circular plate), or is more of its weight concentrated near the edge (like a bicycle wheel)?

hello94857 · Feb 2, 2014, 02:54 PM

Thanks for your answer but I now understand myself, as there is very low friction and because the drag resistance is as it is a solid symmetrically round rotating object, the drag coefficient is 0 and the friction near 0 the answer is less than a 0.001kg of force

ebaines · Feb 2, 2014, 04:18 PM

From the private message you sent me where you indicated that 10 minutes is sufficient to spin the disc up to 2000 RPM, we can now do some calculations:

The governing equation is: $T =I \alpha$ , where T = torque applied, I = the disc's moment of inertia, and $\alpha$ = the disc's angular acceleration.

The angular acceleration is the change of rotational speed divided by time: $\alpha = \frac {\Delta \omega}{\Delta t}$

$\Delta \omega = 2000 \ \text {rev/min}\ \times 2 \pi \text{radian/rev} \ \times \ 1 \text {min}/60 s = 209.4 s^{-2}$

$\Delta T = 10 \ \text {min} \ \times \ 60 \text{s/min} = 600 s$ .

So the acceleration is: $\alpha = \frac {209 s^{-1}}{ 600s} = 0.35 s^{-2}$

The moment of inertia of a disc is I=mr^2 (assuming all the mas is concentrated on the rim), so I = 0.5 Kg-m^2

Hence the torque needed is $0.35 \frac 1 {s^{2}}\ \times \ 0.5 {Kg-m^2} =0.175 N-m$ .

For a constant force applied at 0.75m, the force needed is:

$\frac {0.175 N-m}{0.75 m} = 0.24 N$ .

This is equivalent to the force of gravity on a 23 gram object.