Rotational Motion of a spun disk

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SUMMARY

The discussion focuses on calculating the torque required for a 200 g, 20-cm-diameter plastic disk to accelerate from 0 to 1800 rpm in 4.0 seconds. The moment of inertia (I) is calculated using the formula I = ½ MR², resulting in I = 0.001 kg·m². To find the torque (T), the angular acceleration (α) must first be determined by converting the final angular velocity from rpm to rad/s and applying the kinematic equation. The final torque is expressed as T = Iα.

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  • Understanding of rotational dynamics, specifically torque and moment of inertia.
  • Familiarity with angular velocity and its conversion from rpm to rad/s.
  • Knowledge of kinematic equations for rotational motion.
  • Basic algebra for manipulating equations and solving for unknowns.
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  • Learn how to convert angular velocity from rpm to rad/s.
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  • Explore the relationship between torque, moment of inertia, and angular acceleration.
  • Practice solving problems involving rotational dynamics using various mass and radius values.
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A 200 g, 20-cm-diameter plastic disk is spun on an axle through its
center by an electric motor. What torque must the motor supply to take
the disk from 0 to 1800 rpm in 4.0s? (Given: I = ½ MR2)




Relevant equations
I = 1/2 MR^2
T = Ia


The attempt at a solution
v=0.02m * 1800 rev/min / 4 sec * 60 sec / 1 min * 2 Pi / 1 Rev
v= 10 800 m/s

a = how do i get this?

I = 1/2 (0.2kg)(0.1)^2
I = 0.001

T = Ia
T = (0.001)a
T = answer

is this correct?
 
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What you need to compute is the angular acceleration (\alpha). Just like linear acceleration is defined as change in velocity over time, angular acceleration is change in angular velocity over time. What's the change in angular velocity? (Convert rpm to rad/s.)
 
Use kinematics w = wo + alpha*t. Here wo = zero and w = 2*pi*1800/60. From these values find alpha and then torque.
 

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