Angular Acceleration of a grinding wheel

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SUMMARY

The discussion focuses on calculating the angular acceleration and maximum angular velocity of a vertical grinding wheel with a mass of 62 kg and a radius of 45 cm, using the moment of inertia formula and torque equations. The initial angular acceleration was calculated as 9.31 rad/s², but the correct value is approximately 9.2 rad/s² based on a moment of inertia of 18.38 kg m². For the maximum angular velocity, the participant initially used the incorrect drop height, leading to a calculated value of 3.32 rad/s, which requires adjustment based on the correct height of 0.69 m.

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Homework Statement


A vertical grinding wheel is a uniform disk of mass 62 kg and radius 45 cm. It has a handle of radius 69 cm of negligible mass. A 25-kg load is attached to the handle when it is in the horizontal position.
(a) Neglecting friction, find the initial angular acceleration of the wheel in rad/s^2.
(b) Find the maximum angular velocity of the wheel in rad/s.

Homework Equations


I = 1/2MR^2 + mr^2
t = mgr for torque
a = t/I for angular acceleration
mgr = 1/2Iw^2

The Attempt at a Solution



a) Neglecting friction, find the initial angular acceleration of the wheel.
9.31 rad/s^2
I found I which is 18.18 kg m^2 using the above equation.
I then gound T(t) which is 169.23 Nm
a = t/I = 9.31 rad/s^2

(b) Find the maximum angular velocity of the wheel.
I used mgr = 0.5Iw^2 and got w = 3.32 rad/s

***What am I doing wrong?? I know both are wrong answers. I think I am doing something wrong for part a. Please help!
Thanks
 
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Welcome to PF.

I think for a) your method is OK. I calculate 9.2 m/s2
with I = 18.38 but let's not quibble too much over that.

For b) though I think you are overlooking that the 25 kg weight has dropped .69 m and not the r of .45.
 
I understand it now...Thanks.
 

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