Calculating Angular Acceleration of a Rotating Disk

AI Thread Summary
A disk initially at rest with a reference line at -90° undergoes constant angular acceleration, reaching a final position of 153° after 3.1 seconds. The calculation for angular acceleration initially suggested a value of approximately 50.57, based on the formula theta = theta(0) + omega(0)t + 0.5*alpha*t^2. However, confusion arose regarding the angle measurement, as the problem required radians instead of degrees. The realization that the angles should be converted clarified the misunderstanding, leading to the correct approach for determining angular acceleration. Proper attention to units is crucial in physics calculations.
nahya
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A disk with a radial line painted on it is mounted on an axle perpendicular to it and running through its center. It is initially at rest, with the line at q0 = -90°. The disk then undergoes constant angular acceleration. After accelerating for 3.1 s, the reference line has been moved part way around the circle (in a counterclockwise direction) to qf = 153°.
---
since theta = theta(0) + omega(0)t + 0.5*alpha*t^2, for theta = angle, omega = angular velocity, alpha = angular acceleration, and t in seconds,
153 = -90 + 0*t + 0.5*alpha*t^2
243 = 0.5*alpha*t^2
alpha = 486/t^2
and so alpha should be around 50.5723205... right?!

but it's not!
the degrees seem right, because 243 degrees is the degree of difference. since the disk was initially at rest, omega(0) = 0.
what am i doing wrong?

// got rid of latex cause it wasn't working for some reason
 
Last edited:
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Maybe, the scale goes clockwise so your disk rotates -90 to -180 then 180 to 153, causing a total change of 117 degrees instead of 243. Let me know if that gives you the answer you're looking for...

Regards,
Sam
 
nahya said:
A disk with a radial line painted on it is mounted on an axle perpendicular to it and running through its center. It is initially at rest, with the line at q0 = -90°. The disk then undergoes constant angular acceleration. After accelerating for 3.1 s, the reference line has been moved part way around the circle (in a counterclockwise direction) to qf = 153°.
---
since theta = theta(0) + omega(0)t + 0.5*alpha*t^2, for theta = angle, omega = angular velocity, alpha = angular acceleration, and t in seconds,
153 = -90 + 0*t + 0.5*alpha*t^2
243 = 0.5*alpha*t^2
alpha = 486/t^2
and so alpha should be around 50.5723205... right?!

but it's not!
the degrees seem right, because 243 degrees is the degree of difference. since the disk was initially at rest, omega(0) = 0.
what am i doing wrong?

// got rid of latex cause it wasn't working for some reason

Normally in these problems the angles are expressed in radians, not degrees. The method looks good, anyway.

-Dan
 
I would be surprized if they quoted the angles in degrees and then expected an answer in radians, but its a posibility.

Sam
 
BerryBoy said:
I would be surprized if they quoted the angles in degrees and then expected an answer in radians, but its a posibility.

Sam

Believe me, it's happened before!

-Dan
 
omg... they DID want it in radians.....
hahaha.. I'm sorry. i wasn't careful in reading the problem.
(in the answer box it says rad/s^2. bleh)
 

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