What Torque is Needed to Spin a 250g Disk to 2000 RPM in 4.5 Seconds?

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To calculate the torque required to spin a 250g disk to 2000 RPM in 4.5 seconds, the torque formula T = Iα is used, where I is the moment of inertia and α is the angular acceleration. The moment of inertia for a solid disk is calculated as I = 1/2Mr^2, yielding 0.00125 kg·m² for the given mass and radius. The angular acceleration is determined by converting 2000 RPM to radians per second and dividing by the time, resulting in approximately 46.54 rad/s². The torque is then calculated as T = Iα, leading to a value of 0.06 Nm. The original poster is seeking clarification on the correctness of their calculations.
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Homework Statement


A 250 g , 20.0-cm-diameter plastic disk is spun on an axle through its center by an electric motor.

Homework Equations


What torque must the motor supply to take the disk from 0 to 2000 rpm in 4.50 s?

The Attempt at a Solution


I know that Torque = Inertia times angular acceleration or T = Ia.
I tried using the formula for inertia of a solid disk, I = 1/2Mr^2. So with that I get 1/2(.25)(.1^2) = .00125. Then i found angular acceleration: 2000 rpm = 209.44 rad/s divided by 4.5 s so 209.44 rad/s / 4.5s = 46.54 s^2. I multiply these two number together and get .00125*46.54 = .06 Nm. However, this is wrong. What am I doing wrong?
 
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Your working seems correct to me...
 

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