A disk with mass m = 11.5 kg and radius R = 0.3 m begins at rest and accelerates

Question

A disk with mass m = 11.5 kg and radius R = 0.3 m begins at rest and accelerates uniformly for t = 16.6 s, to a final angular speed of omega = 28 rad/s. What is the angular acceleration of the disk? rad/s^2 What is the angular displacement over the 16.6 s? rad What is the moment of inertia of the disk? kg-m^2 What is the change in rotational energy of the disk? J What is the tangential component of the acceleration of a point on the rim of the disk when the disk has accelerated to half its final angular speed? m/s^2 What is the magnitude of the radial component of the acceleration of a point on the rim of the disk when the disk has accelerated to half its final angular speed? m/s^2 What is the final speed of a point on the disk half-way between the center of the disk and the rim? m/s What is the total distance a point on the rim of the disk travels during the 16.6 seconds? m

Explanation / Answer

5)

= / t

= (f - o)/t

   = (28 rad/s- 0)/(16.6 s)

    =1.686 rad/s^2

at= * R

    = 1.686 * 0.3 m

    = 0.5 m/s^2

6)

ar = ² r

     = ((28/2) rad/s)² * 0.3 m

     = 58.8 m/s²

7)

v = r

     = 28 rad/s * (1/2)*0.3 m

     = 4.2 m/s

8)

theta=1/2 t2

        =1/2*1.686 rad/s2*(16.6 s)2

        =232 rads

. s = theta* r

     = 232 rads * 0.3 m

     =69.6 m

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