A disk with mass m = 9.3 kg and radius P. = 0.39 m begins at rest and accelerate

Question

A disk with mass m = 9.3 kg and radius P. = 0.39 m begins at rest and accelerates uniformly for t = 17.6 s, to a final. angular speed of omega = 26 rad/s. 1) What is the angular acceleration of the disk? rad/s^2 2) What is the angular displacement over the 17.6 s? rad 3) What is the moment of inertia of the disk? kg middot m^2 4) What is the change in rotational energy of the disk? J 5) 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 6) 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 7) What is the final speed of a point on the disk half-way between the center of the disk and the rim? m/s 8) What is the total distance a point on the rim of the disk travels during the 17.6 seconds? m

Explanation / Answer

5.) Acceleration is uniform.So, it will remain constant through out the motion.

a=change in angular velocity/time taken = 26/17.6=1.49 rad/s^2

Tangential componet of acceleration= ra=0.39*1.49=0.58 m/s^2

6.) Angular speed= 13 rad/s

Radial accleration= rw^2=0.39(13)^2=65.91 m/s^2

7.) Final speed= rw=(0.39/2)(26)=10.14 m/s

8.) Angular distance,l= ut+1/2at^2

Since, u=0

l=1/2 (1.49)(17.6)^2=230.7 rad

distance = r (angular distance)=230.7(0.39)=89.93 m

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