Question

A wheel 1.55 m in diameter lies in a vertical plane and rotates about its central axis with a constant angular acceleration af 3.55a2. The wheel stats at reat at t-o, arnd the radius vector of a certaln pcirt P on the rim makes an angle of 57.3e with tha horizontal at this time. At t - 2.00 s, firnd the following (a) the angular speed of the wheel rad/s (b) the tangential speed of the paint ρ ]V m/s 55025 (c) the total acceleration of the point P magnitude 30.484x Your response differs from the correct answer by more than 10%. Double cneck your calculabans. Youresponse difters from thertswer by r calculatians. with respect to the radius to point P direction 66 swer by more thה10-%, nouble check you Your response differs from the correct ar (d) the angular position of the point P 104.099X Your response differs significantly from the correct answe. Rework your solutin from the k your solution from the beginning and check eacn step canefully. red Need Help? Master

Answer 1

answer)b) here for total acceleration

a= $\sqrt{}$ tang a² + radial a²

tang a=3.55 rad/s²*0.775=2.7513m/s²

radial a=v²/r=5.5025²/0.775=39.06775 m/s²

a= $\sqrt{}$ 2.7513²+39.06775²= $\sqrt{}$ 1533.8585=39.2 m/s²

magnitude=39.2 m/s² or 39.23 m/s² or 39.16m/s²

direction=8.1rad=464.096⁰-360^o=104⁰

so direction=104^o or 104.1^o or 104.096^o

c) the angular position for point P

from kinematics we have

s=ut+1/2at²

s=1/2at²=1/2*3.55*2²=7.1rad

we know that 57.3^o=1 rad

so total total displacement=7.1rad+1rad=8.1 rad

so the answer is 8.1 rad or 8.10 rad