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During the time a compact disc (CD) accelerates from rest to a constant rotational speed of 477 rev/min, it rotates through an angular displacement of 0.250 rev. What is the angular acceleration of the CD

Respuesta :

gilcarusolautaro

Answer:

The angular acceleration of the CD is α= 15.9 rad/s².

Explanation:

wf= 477 r/min = 49.95 rad/sec

wi= 0 rad/sec

θ= 0.25rev = 1.57 rad

θ= (wf²-wi²) / 2 * α

α= 15.9 rad/sec²

xero099

Answer:

[tex]\alpha = 455058\,\frac{rev}{min^{2}}[/tex]

Explanation:

Let consider that CD accelerates at constant rate. Then, angular acceleration experimented by the CD is:

[tex]\alpha = \frac{(477\,\frac{rev}{min} )^{2}}{2\cdot (0.250\,rev)}[/tex]

[tex]\alpha = 455058\,\frac{rev}{min^{2}}[/tex]

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