# Calculate Torque Required

I am working on a school project that requires me to build a machine. I have designed the machine but I am stuck at torque calculation. The problem goes like this, the machine supposes to rotate a stack of trays (5 trays stacking up vertically) at a constant speed.

Below is the information:

Mass of each tray: 8 kg

Tray Radius: 0.3 m

Moment of inertia of each tray: 0.333 kg m^2 Let's just say the friction is very small and can be neglected. The rotational speed should reach 1200 rev/min in 5 seconds. How do I calculate the torque required to rotate the stack of trays?

Any help would be appreciated!

• Torque will accelerate the trays until the friction of rotation equals the applied torque. If you don't have any rotational losses, any torque large enough to put the trays in motion will make them move. The data shown goes better with something like "how much torque to get the platter to spin at 10 rev/min within 15 seconds." – Tiger Guy May 21 '20 at 1:56
• Let’s assume the friction is very small so it can be neglected and the machine needs to reach 1200 rev/min within 5 seconds. How much torque required to rotate it? Thanks! – lalala May 21 '20 at 2:46

## 1 Answer

If we follow your assumption, ignoring friction.

We calculate for a Torque to produce an angular acceleration of 1200/5 per second.

$$T=\alpha L \ = \\1200/5*2\pi*l= 753.98*0.333*5*9.8=12,302.7Nm$$

• 1200 is in RPM. Don't you need to convert that to rad/sec? (You converted to radians, but not seconds). Where does the number 753.98 come from? (I think you missed the 2 in 2*pi.) What is the factor 9.8? (It is not the gravitational value since that is not needed. kgm^2/s^2 = Nm.) With these changes, I am getting 41.8 N*m torque required. – JohnHoltz May 22 '20 at 22:08
• @JohnHoltz, right. my apologies. I forgot to divide by seconds. The cost of hasty answering on the phone. later I fix the arithmetic. – kamran May 23 '20 at 4:49
• That is a lot of torque. – StainlessSteelRat Jul 16 '20 at 22:30