Question

10) At a time t = 2.90 s , a point on the rim of a wheel with a radius of 0.200 m has a tangential speed of 45.0 m/s as the wheel slows down with a tangential acceleration of constant magnitude 10.3 m/s2 .

Part A

Calculate the wheel's constant angular acceleration.]

Part B

Calculate the angular velocity at t = 2.90 s .

Part C

Calculate the angular velocity at t=0.

Part D

Through what angle did the wheel turn between t=0 and t = 2.90 s ?

Part E

Prior to the wheel coming to rest, at what time will the radial acceleration at a point on the rim equal g = 9.81 m/s2 ?

Answer 1

a.)

angular accel in rad/s^2 = tangential accel / r = 10.3m/s² / 0.2 m = 51.5 rad/s²

b.)

angular velocity in rad/s = tangential speed / r = 45.0m/s / 0.2 m = 225 rad/s

c.)

accel is negative (slowing down)
so velocity at t=0 = u = v + a*t = 225 + (51.5)*2.9 rad/s = 374.35 rad/s

d.)

angle s = u*t - 0.5*a*t²
s = (374.35*2.9) - 0.5*51.5*(2.9)² = 1085.615 rad - 216.5575 rad = 869.0575 radians

e.)

radial accel = v²/r = 9.81m/s²
v = sqrt(9.81*0.2) = 1.50 m/s = 1.40 / 0.2 rad/s = 7 rad/s
This occurs when t = (u - v) / a = (374.35 - 7) / -51.5 = 7.13 s