Angular acceleration, is my method correct?

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SUMMARY

The discussion centers on calculating the angular acceleration and linear acceleration of a pulley with mass M=4 kg and radius R=0.4 m, modeled as a uniform solid disk with a moment of inertia I=0.5MR². The angular speed reached after 6 seconds is T=60 rad/s. The correct angular acceleration is determined to be 10 rad/s², while the tangential acceleration is calculated using the formula a = r * α, resulting in 4 m/s². The initial calculations presented were incorrect, and the correct method was clarified by another participant.

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Homework Statement



A Pulley has mass M=4 kg and radius R=0.4 m. Assume that it is a uniform solid disk so that its moment of inertia is I= .5MR^2. A massless cord is wrapped around it and a tension force Ft is applied. The pulley starts from rest. After the tension force has been applied for 6 seconds, the angular speed has reached T=60 rads/s.


Homework Equations



a.) What is the pulley's angular acceleration?

b.) What is the linear (tangential) acceleration of a point on the rim of the pulley?


The Attempt at a Solution



a.) V=(2*pi*R)/T
V= .418 m/s

w=v/r
w=1.045 rads/sec

angular acceleration= w2-w1/t2-t1
= (1.045-0)/(6-0)
= .15 rads/s^2

b.) Tangential acceleration= r*alpha
=.o2 m/s^2


Can anyone please tell me if this is correct?? Thank you thank you!
 
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Consider that it is uniform acceleration and

ω = a*t

a) is given simply by

60 = a*6 or a = 10 rad/sec2

b) would be determined by the same method you used.
 

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