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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).:

Isometric view

Dimensions are in cm: enter image description hereFront view 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: enter image description here 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):

Isometric view

Dimensions are in cm: Front view

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
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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: enter image description here 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: enter image description here 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: enter image description here 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
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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: enter image description hereenter image description here 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: enter image description here 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: enter image description here 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.

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