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 4 added 103 characters in body edited Oct 19 '18 at 11:08 Mark Booth♦ 3,64611 gold badge1919 silver badges4646 bronze badges I am trying to spec out a motor's torque required to rotate a platform about an axis. The attached This diagram makes it better to understand(it's like a rotisserie).: Dimensions are in cm: I I have an arbitrary shaped object on the platform that I simplified to a cuboid to calculate the moment of inertia. The The combined weight is about 25kg. I I use the moment of inertia formula for a cuboid along the length axis and use the parallel axis theorem to move the axis to where I have the rod, I get: I = (1/2) * m * (w^2 + h^2) + m*((0.5*h)^2) = (1/2) * 25kg * (0.4^2+ 0.5^2) + 25 * (0.2^2) = 6.125 kg-m^2  Assuming I want to reach 5 rpm in 5 seconds. I have  5rpm = (2 * pi * 5)/60 rad/s alpha = ((2 * pi * 5)/60 - 0)/(5-0) rad/s^2 = pi/30 rad/s^2  T = I * alpha = 6.125 * (pi/30) = 0.64N-m Now I am not entirely sure if this calculation is correct. I had a 5N-m rated dc motor lying around and I fit it to the platform. The motor was able to rotate the platform about 45 degree clockwise but was not able to come back to zero degrees. Am I missing something in the above calculation? Gravity doesn't feature in my equations. There There could be other factors like friction, or the gearbox in the motor? I am trying to spec out a motor's torque required to rotate a platform about an axis. The attached diagram makes it better to understand(it's like a rotisserie). Dimensions are in cm: I have an arbitrary shaped object on the platform that I simplified to a cuboid to calculate the moment of inertia. The combined weight is about 25kg. I use the moment of inertia formula for a cuboid along the length axis and use the parallel axis theorem to move the axis to where I have the rod, I get: I = (1/2) * m * (w^2 + h^2) + m*((0.5*h)^2) = (1/2) * 25kg * (0.4^2+ 0.5^2) + 25 * (0.2^2) = 6.125 kg-m^2  Assuming I want to reach 5 rpm in 5 seconds. I have  5rpm = (2 * pi * 5)/60 rad/s alpha = ((2 * pi * 5)/60 - 0)/(5-0) rad/s^2 = pi/30 rad/s^2  T = I * alpha = 6.125 * (pi/30) = 0.64N-m Now I am not entirely sure if this calculation is correct. I had a 5N-m rated dc motor lying around and I fit it to the platform. The motor was able to rotate the platform about 45 degree clockwise but was not able to come back to zero degrees. Am I missing something in the above calculation? Gravity doesn't feature in my equations. There could be other factors like friction, or the gearbox in the motor? I am trying to spec out a motor's torque required to rotate a platform about an axis. This diagram makes it better to understand(it's like a rotisserie): Dimensions are in cm: I have an arbitrary shaped object on the platform that I simplified to a cuboid to calculate the moment of inertia. The combined weight is about 25kg. I use the moment of inertia formula for a cuboid along the length axis and use the parallel axis theorem to move the axis to where I have the rod, I get: I = (1/2) * m * (w^2 + h^2) + m*((0.5*h)^2) = (1/2) * 25kg * (0.4^2+ 0.5^2) + 25 * (0.2^2) = 6.125 kg-m^2  Assuming I want to reach 5 rpm in 5 seconds. I have  5rpm = (2 * pi * 5)/60 rad/s alpha = ((2 * pi * 5)/60 - 0)/(5-0) rad/s^2 = pi/30 rad/s^2  T = I * alpha = 6.125 * (pi/30) = 0.64N-m Now I am not entirely sure if this calculation is correct. I had a 5N-m rated dc motor lying around and I fit it to the platform. The motor was able to rotate the platform about 45 degree clockwise but was not able to come back to zero degrees. Am I missing something in the above calculation? Gravity doesn't feature in my equations. There could be other factors like friction, or the gearbox in the motor? 3 better diagram, update calculation edited Oct 19 '18 at 0:59 rookie 16366 bronze badges I am trying to spec out a motor's torque required to rotate a platform about an axis. The attached diagram makes it better to understand(it's like a rotisserie). Dimensions are in cm: I have an arbitrary shaped object on the platform that I simplified to a cuboid to calculate the moment of inertia. The combined weight is about 25kg. I use the moment of inertia formula for a cuboid along the length axis and use the parallel axis theorem to move the axis to where I have the rod, I get: I = (1/2) * m * (w^2 + h^2) + m*((0.5*h)^2) = (1/2) * 25kg * (0.4^2+ 0.5^2) + 25 * (0.2^2) = 156.125 kg-m^2  Assuming I want to reach 5 rpm in 5 seconds. I have  5rpm = (2 * pi * 5)/60 rad/s alpha = ((2 * pi * 5)/60 - 0)/(5-0) rad/s^2 = pi/30 rad/s^2  T = I * alpha = 156.125 * (pi/30) = 10.58N64N-m Now I am not entirely sure if this calculation is correct. I had a 5N-m rated dc motor lying around andand I fit it to the platform. The motor was able to go downrotate the platform about 45 degree clockwise but was not able to lift the weightcome back to zero degrees. Am I missing something in the above calculation? Gravity doesn't feature in my equations. There could be other factors like friction, or the gearbox in the motor.? I am trying to spec out a motor's torque required to rotate a platform about an axis. The attached diagram makes it better to understand(it's like a rotisserie). Dimensions are in cm: I have an arbitrary shaped object on the platform that I simplified to a cuboid to calculate the moment of inertia. The combined weight is about 25kg. I use the moment of inertia formula for a cuboid along the length axis and use the parallel axis theorem to move the axis to where I have the rod, I get: I = (1/2) * m * (w^2 + h^2) + m*((0.5*h)^2) = (1/2) * 25kg * (0.4^2+ 0.5^2) = 15.125 kg-m^2  Assuming I want to reach 5 rpm in 5 seconds. I have  5rpm = (2 * pi * 5)/60 rad/s alpha = ((2 * pi * 5)/60 - 0)/(5-0) rad/s^2 = pi/30 rad/s^2  T = I * alpha = 15.125 * (pi/30) = 1.58N-m Now I am not entirely sure if this calculation is correct. I had a 5N-m rated dc motor lying around and I fit it to the platform. The motor was able to go down about 45 degree clockwise but was not able to lift the weight back to zero degrees. Am I missing something in the above calculation? Gravity doesn't feature in my equations. There could be other factors like friction, or the gearbox in the motor. I am trying to spec out a motor's torque required to rotate a platform about an axis. The attached diagram makes it better to understand(it's like a rotisserie). Dimensions are in cm: I have an arbitrary shaped object on the platform that I simplified to a cuboid to calculate the moment of inertia. The combined weight is about 25kg. I use the moment of inertia formula for a cuboid along the length axis and use the parallel axis theorem to move the axis to where I have the rod, I get: I = (1/2) * m * (w^2 + h^2) + m*((0.5*h)^2) = (1/2) * 25kg * (0.4^2+ 0.5^2) + 25 * (0.2^2) = 6.125 kg-m^2  Assuming I want to reach 5 rpm in 5 seconds. I have  5rpm = (2 * pi * 5)/60 rad/s alpha = ((2 * pi * 5)/60 - 0)/(5-0) rad/s^2 = pi/30 rad/s^2  T = I * alpha = 6.125 * (pi/30) = 0.64N-m Now I am not entirely sure if this calculation is correct. I had a 5N-m rated dc motor lying around and I fit it to the platform. The motor was able to rotate the platform about 45 degree clockwise but was not able to come back to zero degrees. Am I missing something in the above calculation? Gravity doesn't feature in my equations. There could be other factors like friction, or the gearbox in the motor? 2 better diagram edited Oct 19 '18 at 0:51 rookie 16366 bronze badges I am trying to spec out a motor's torque required to rotate a platform about an axis. The attached diagram makes it better to understand(it's like a rotisserie). Dimensions are in cm: I have an arbitrary shaped object on the platform that I simplified to a cuboid to calculate the moment of inertia. The combined weight is about 25kg. I use the moment of inertia formula for a cuboid aroundalong the length axis and use the parallel axis theorem to move the axis to where I have the rod, I get: I = (1/2) * m * (w^2 + h^2) + m*((0.5*h)^2) = (1/2) * 25kg * (0.4^2+ 0.5^2) = 15.125 kg-m^2  Assuming I want to reach 5 rpm in 5 seconds. I have  5rpm = (2 * pi * 5)/60 rad/s alpha = ((2 * pi * 5)/60 - 0)/(5-0) rad/s^2 = pi/30 rad/s^2  T = I * alpha = 15.125 * (pi/30) = 1.58N-m Now I am not entirely sure if this calculation is correct. I had a 5N-m rated dc motor lying around and I fit it to the platform. The motor was able to go down about 45 degree clockwise but was not able to lift the weight back to zero degrees. Am I missing something in the above calculation? Gravity doesn't feature in my equations. There could be other factors like friction, or the gearbox in the motor. I am trying to spec out a motor's torque required to rotate a platform about an axis. The attached diagram makes it better to understand. Dimensions are in cm: I have an arbitrary shaped object on the platform that I simplified to a cuboid to calculate the moment of inertia. The combined weight is about 25kg. I use the moment of inertia formula for a cuboid around the length axis and use the parallel axis theorem to move the axis to where I have the rod, I get: I = (1/2) * m * (w^2 + h^2) + m*((0.5*h)^2) = (1/2) * 25kg * (0.4^2+ 0.5^2) = 15.125 kg-m^2  Assuming I want to reach 5 rpm in 5 seconds. I have  5rpm = (2 * pi * 5)/60 rad/s alpha = ((2 * pi * 5)/60 - 0)/(5-0) rad/s^2 = pi/30 rad/s^2  T = I * alpha = 15.125 * (pi/30) = 1.58N-m Now I am not entirely sure if this calculation is correct. I had a 5N-m rated dc motor lying around and I fit it to the platform. The motor was able to go down about 45 degree clockwise but was not able to lift the weight back to zero degrees. Am I missing something in the above calculation? Gravity doesn't feature in my equations. There could be other factors like friction, or the gearbox in the motor. I am trying to spec out a motor's torque required to rotate a platform about an axis. The attached diagram makes it better to understand(it's like a rotisserie). Dimensions are in cm: I have an arbitrary shaped object on the platform that I simplified to a cuboid to calculate the moment of inertia. The combined weight is about 25kg. I use the moment of inertia formula for a cuboid along the length axis and use the parallel axis theorem to move the axis to where I have the rod, I get: I = (1/2) * m * (w^2 + h^2) + m*((0.5*h)^2) = (1/2) * 25kg * (0.4^2+ 0.5^2) = 15.125 kg-m^2  Assuming I want to reach 5 rpm in 5 seconds. I have  5rpm = (2 * pi * 5)/60 rad/s alpha = ((2 * pi * 5)/60 - 0)/(5-0) rad/s^2 = pi/30 rad/s^2  T = I * alpha = 15.125 * (pi/30) = 1.58N-m Now I am not entirely sure if this calculation is correct. I had a 5N-m rated dc motor lying around and I fit it to the platform. The motor was able to go down about 45 degree clockwise but was not able to lift the weight back to zero degrees. Am I missing something in the above calculation? Gravity doesn't feature in my equations. There could be other factors like friction, or the gearbox in the motor. 1 asked Oct 18 '18 at 20:41 rookie 16366 bronze badges