Problem 16-1: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)

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1. We are given the angular velocity as a function of time, the radius of the disk, and the time of interest. To find the velocity and acceleration of point A on the rim, we need omega, alpha, and the radius. We find alpha by differentiating omega with respect to time.
2. Here is the kinematic diagram of the disk. Point A sits on the rim. The tangential velocity and tangential acceleration point perpendicular to the radius, shown in green. The normal acceleration points from A straight toward the center O, shown in orange.
3. Here is the kinematic diagram of the disk. Point A sits on the rim. The tangential velocity points perpendicular to the radius, the tangential acceleration points in the same tangential direction, and the normal acceleration points from A straight toward the center O.
4. Here is the kinematic diagram of the disk. Center O is marked at the middle. Point A sits on the rim at the end of the radius. The tangential velocity and tangential acceleration point perpendicular to the radius, shown in green. The normal acceleration points from A straight toward the center O, shown in orange. The curved arrow at O indicates the direction of angular velocity omega.
5. Substituting t equals 0.5 seconds into the angular velocity formula: 5 times 0.5 squared is 1.25, plus 2 gives omega equals 3.25 radians per second.
6. Differentiating the angular velocity with respect to time gives alpha equals 10 t. At t equals 0.5 seconds, alpha equals 5.0 radians per second squared.
7. The speed of point A is the radius times omega: 0.8 meters times 3.25 radians per second equals 2.60 meters per second.
8. The tangential acceleration is radius times alpha: 0.8 times 5.0 equals 4.0 meters per second squared. The normal acceleration is radius times omega squared: 0.8 times 3.25 squared equals 0.8 times 10.5625, which gives 8.45 meters per second squared, directed toward the center.
9. Since the tangential and normal accelerations are perpendicular to each other, the total acceleration magnitude is the square root of their squares added together: square root of 16 plus 71.40 equals square root of 87.40, which gives approximately 9.35 meters per second squared.
10. To summarize: at t equals 0.5 seconds, the angular velocity is 3.25 radians per second and the angular acceleration is 5.0 radians per second squared. Point A on the rim has a speed of 2.60 meters per second, a tangential acceleration of 4.0 meters per second squared, a normal acceleration of 8.45 meters per second squared, and a total acceleration magnitude of approximately 9.35 meters per second squared.