Question

(b) What is the angelased the wheels when the hard the total distance! Need Help? MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER 4. (-/1 Points) DETAILS SERCP11 7.P.012. A 4.0 cm diameter disk rotates with a constant angular acceleration of raw from restatt-, and a line drawn from the center of the deck to a point on them the des make an angle 57.3" with the positive and at the time Att 2.45 , find the angular speed of the wheel (6) ALT=2.48, in the magnitude of the new velocity and tangential acceleration of P na velocity ms tangential acceleration (C) AT 2.48 s, find the position of P (in degrees, with respect to the positive axis). counterclockwise from the Need Help? 40 90 6 7 5 8 9 o 2 Р T E W Q R Y

Answer 1

Given of a * Diameter disk (d) = 44cm = 0.40m * Angular acceleration (Q): 2-80 vad/s2 initial angulor velocity (29) = 0 * at inihal * Initial o = 57.3° with ave a quis * 1573 time til 44cm +x we know Find @ At t = 2:485, angular speed of wheel. W = Witat so final angular speed (wp] = 0+ 2.8 X 2.48 Wf = 6. guy rad/s

6 At t=2.485, find and so, 1-52.7 m/s 1 Angle measured counter-clockwise with tx arisa magnitude of lineas velocity tangential acceleration at p. Relation between linear ang and angular Velocity VE 92.69, where r is radius of wheel >V u angulae speed ww Linear velocity at point Ply)= 92.00 0.22 x Gogun Vp At +=0, Wi=o so Ve=0 At t = 2.485, Wq = 6.944, and up = 1527 using formula V- utat Y = Vitat 1.527 = 0 + 0 x 2.48 a = 1.527 2.48 a 0.615 m/s2 At 2.485, find position of P Angular displacement at 2.485 is given as Wit tat2 eP 0x++ { x 2.8862.48)? 573 +816) 65.91 ce 2 - O = 8.61° w ta