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determine the torque

3/14/2012 10:29 AM	Rate (0)
Mothilal Posts: 165 Rating: (25)	hello experts could you please help me to find the following terms? statement: I have a circular metal (say iron) disc with a mass of 100kg and the Its diameter is 1meter.The disc is connectedwith a motor shaft (shaft dia is 250 cm). The disc is drivenby motor shaft directly. a) ifwe want to rotate the disc to 1500 rpm within 10 sec, how much acceleration torque is required? (starting torque) b) torque required to maintain the disc at 1500 rpm (running torque) c) torque required to stop the disc within 10 sec (braking torque) d) Moment of inertia of the disc If you have any idea about this , please let me know... Thanks Regards Mothilal
Mothilal Posts: 165 Rating: (25)
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4/5/2012 1:13 PM	Rate (2)
Technology Team Platinum Expert Joined: 10/11/2006 Last visit: 12/22/2025 Posts: 6062 Rating: (828)	Hello, all this questions any even much more of technical data for your applicationcan be calculated with the Siemens tool SIZER: https://support.automation.siemens.com/WW/llisapi.dll?func=cslib.csinfo&lang=de&objid=10804987&subtype=133100&caller=view (=download link) Figure it out, it is really helpful!
	With best regards, Technology Team Application consulting and pre sales support for SINAMICS, SIMATIC Technology, SIMOTION, SIMOTICS and MOTION CONNECT
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This contribution was helpful to 1 thankful Users Mothilal

4/26/2012 7:52 PM	Rate (2)
Deleted User Posts: 767 Rating: (69)	Hello a. starting Torque 196.25 Nm b. theoretically since a disc does not inflict any friction, running torque would be practically zero. c. braking torque 196.25 Nm d. moment of inertia of disc J(disc) = 12.5 kgmm J(disc) = (1/2) * mass * radius * radius = 1/2 * 100kg * 0.5m * 0.5m = 12.5 kgmm 1500 rev/min -> 25 rev/sec alpha = w/time = 25 rev/sec / 10 sec = 2.5 rev/secsec Torque acceleration = J * alpha * 2 * Pi = 12.5 kgmm* 2.5 rev/secsec* 2 * 3.14 = 196.25 Nm
Deleted User Posts: 767 Rating: (69)
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