A disk of of radius 237cm rotates about it\'s axis. Points on the disks rim unde
ID: 2011512 • Letter: A
Question
A disk of of radius 237cm rotates about it's axis. Points on the disks rim undergo tangential acceleration of magnitude 2.31 m/s2. at a particular time the rim has a tangential speed of 1.05 m/s. At a time 0.993 seconds later, what is the tangential speed, v, of a point on the rim, the magnitude of the points radial acceleration, ar and a magnitude of it's total acceleration atot?
V=_________
ar=_________
atot__________
please show the steps and any equations used to arrive at answers. I want to learn how to do them.
Explanation / Answer
Data: Radius, r = 237 cm = 2.37 m Tangential acceleration, at = 2.31 m/s^2 Tangential speed, vo = 1.05 m/s Time, t = 0.993 s Solution: (a) v = vo + ( at * t ) = 1.05 + ( 2.31 * 0.993 ) = 1.05 + 2.29 = 3.34 m/s Ans: Tangential speed after 't' s = 3.34 m/s (b) Radial acceleration, ar = v^2 / r = 3.34^2 / 2.37 = 4.72 m/s^2 Ans: Radial acceleration, ar = 4.72 m/s^2 (c) Total acceleration, at = [ ( ar )^2 + ( at )^2 ] = [ ( 4.72 )^2 + ( 2.31 )^2 ] = 5.25 m/s^2 Ans: Total acceleration, at = 5.25 m/s^2Related Questions
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