Determine the magnitude of the angular acceleration of the discus

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The discus thrower accelerates the discus from rest to 25 m/s while completing 1.25 revolutions on a 1m radius arc. The initial calculations for angular velocity yield 25 rad/s. The total distance moved is calculated as approximately 7.854 m, leading to an acceleration of 39.79 m/s². The angular acceleration is then determined to be 39.79 rad/s². A correction is noted regarding the use of 1.5 instead of 1.25 in the calculations, which affects the final result.
UrbanXrisis
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A discus thrower accelerates a discus from rest to a speed of 25 m/s by whirling it though 1.25 rev. Assume the discus moves on the arc of a circle 1m in radius.

Determine the magnitude of the angular acceleration of the discus:

ω=angular velocity

ω=v/r=25m/s / 1m = 25rad/s

ω^2=2ad
(25rad/s)^2=2a1.5(pi2)
a=33m/s^2

is this correct?
 
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UrbanXrisis said:
A discus thrower accelerates a discus from rest to a speed of 25 m/s by whirling it though 1.25 rev. Assume the discus moves on the arc of a circle 1m in radius.

Determine the magnitude of the angular acceleration of the discus:

ω=angular velocity

ω=v/r=25m/s / 1m = 25rad/s

ω^2=2ad
(25rad/s)^2=2a1.5(pi2)
a=33m/s^2

is this correct?
{Total Distance Moved} = d = (1.25 rev)*(2*Pi*r) = (1.25 rev)*{2*Pi*(1 m)} = 7.854 m
{Acceleration} = (v^2)/(2*d) = {(25 m/sec)^2}/{2*(7.854 m)} = 39.79 m/sec^2
{Angular Acceleration} = {Acceleration}/r = {39.79 m/sec^2}/(1 m)
{Angular Acceleration} = 39.79 rad/sec^2


~~
 
UrbanXrisis said:
ω=v/r=25m/s / 1m = 25rad/s

ω^2=2ad
(25rad/s)^2=2a1.5(pi2)
a=33m/s^2

is this correct?
Your method is correct but you plugged in 1.5 instead of 1.25. Correct that error and you'll get the answer that xanthym posted.
 
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