Question

7P	A	n	a	r
The position of a block moving outward along the slot in the platform is r = 0.022 t^4 m/s where r is in m and t is in seconds. The platform angular velocity increases at a constant rate of 0.5 rad/s/s. If the block and platform start from rest at r = 0 (the center), determine the magnitudes of the block (a) velocity and (b) acceleration when the block is at position r = 8 m.	0.022	4	0.5	8

Answer 1

Position of the block is = 0.0227 ms angular acceleration is 2 = 0.5 rad. /?. we calculate it when now = 8 m. 8 = 0.022 ++ 0.022 it = (363.6363) / 4.366 seconds. t = from first law of vrotation now w = Wot het

where wo = o rad guc. (initial angular speed) W = 0 + 0.5 x 4.366 w = 2.183 vtad /sec. fourmala Here we use the radius & angular velocity = Linear relocity o = vw 8 X 2.183 va 17.464 m) (b) of acceleration the belock is Linear 9 JTD 9 = 8x0.5 9 = 4 m/?

of acculeration the block is radial 17. qe = (17.464) 9 = 38.123 m/s² acceleration is magnitude of now a = 9,+ qa? ( 38 a = 4 (4) + ( 38 1238 9 38-3323 2