A disk of radius 0.30 m and mass 7 kg is spinning clockwise at22 rad/s about a f

Question

A disk of radius 0.30 m and mass 7 kg is spinning clockwise at22 rad/s about a fixed axis in the center. After rotatingthrough 65 rad the disk is spinning in the opposite direction at 18rad/s. a) At what time does the disk momentarily come to rest? b) What is the smallest magnitude the total acceleration canhave as the disk rotates through the 65 radians? Describe thevelocity of the wheel when this happens. c) What will the time be the first time the radialacceleration is momentarily g (i.e. the shortest time to when theradial acceleration is g)? A disk of radius 0.30 m and mass 7 kg is spinning clockwise at22 rad/s about a fixed axis in the center. After rotatingthrough 65 rad the disk is spinning in the opposite direction at 18rad/s. a) At what time does the disk momentarily come to rest? b) What is the smallest magnitude the total acceleration canhave as the disk rotates through the 65 radians? Describe thevelocity of the wheel when this happens. c) What will the time be the first time the radialacceleration is momentarily g (i.e. the shortest time to when theradial acceleration is g)?

Explanation / Answer

We know that =0t +0.5t2   and = 0 - t then t = (0 - )/ keep inmind that 0 and are apposing. substituting in to we have =0t +0.5t2   =0(0 -)/ + 0.5((0 -)/ )2   =0(0 -)/ + 0. 5(0 -)2/ then is =   [0(0 -)/ + 0. 5(0 -)2 ]/ [note again keep in mind that0 and are apposing ] wehave =  [0(0 + )/ + 0. 5(0 + )2 ]/ a) now we can compute t for =0 using the equation = 0 - t where t=0 / b) as derived yearlier =  [0(0 + )/ + 0. 5(0 + )2 ]/ c) since a=V2/R a= g we have V= ( gR) also V= R we have R = ( gR) = (g/R) and since = 0 - t wecan compute t again as t = ( - 0)/    substituting = (g/R) wehave t=  ((g/R) - 0)/ Just plug and play

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