Problem 16-2: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)
Engineering Mechanics Dynamics · R.C. Hibbeler · Chapter 16
Step-by-step video solution
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Problem and lesson overview
The angular acceleration of a disk is defined by rad/s², where is in seconds. The disk is originally rotating at rad/s. Determine the magnitude of the velocity and the normal and tangential components of acceleration of point A on the disk when s. Point A is located m from the center of the disk.
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- We integrate the angular acceleration to get angular velocity. The integrand is 3 t squared plus 12. Integrating gives t cubed plus 12t plus a constant. Applying the initial condition that omega equals 12 radians per second at t equals zero, we find the constant equals 12. So omega of t equals t cubed plus 12t plus 12.
- Here is the setup. The disk rotates about its fixed center O. Point A sits on the disk, 0.5 meters from the center. At point A, the velocity is tangential — perpendicular to the radius. The normal acceleration points inward toward O, and the tangential acceleration is perpendicular to the radius in the direction of angular acceleration.
- Substituting t equals 2 seconds: omega equals 2 cubed plus 12 times 2 plus 12, which is 8 plus 24 plus 12, giving 44 radians per second.
- We substitute t equals 2 into the angular acceleration formula. 3 times 4 is 12, plus 12 gives 24 radians per second squared.
- The speed of point A equals the radius times the angular velocity. That is 0.5 meters times 44 radians per second, giving a velocity of 22 meters per second.
- The normal component of acceleration is r times omega squared. That is 0.5 times 44 squared. 44 squared is 1936, so the normal acceleration is 968 meters per second squared, directed toward the center.
- The tangential component of acceleration equals r times alpha. That is 0.5 times 24, giving 12 meters per second squared in the tangential direction.
- The total acceleration is the square root of the sum of the squares of the normal and tangential components. The square root of 968 squared plus 12 squared equals approximately 968 meters per second squared. The normal component completely dominates here.
- To summarize: at t equals 2 seconds, point A has a speed of 22 meters per second. Its normal acceleration is 968 meters per second squared directed toward the center, and its tangential acceleration is 12 meters per second squared. The total acceleration magnitude is approximately 968 meters per second squared.
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