Answer:
1.962 rad / s², 1.6 s
Explanation:
Radius of the part = 1.0m / 2 = 0.5 m
angular speed = 30 rpm = 30 rpm × (2πrad / rev) × 1 minutes / 60 seconds = 3.142 rads⁻¹
μk = frictional force / normal ( mg )
normal is the force acting upward against the force of gravity
frictional force = - μk mg
since the body came to rest then
Fnet + Ff = 0
Fnet = - Ff
Fnet = ma
ma = - μk mg
a = - μkg where g = 9.81
a = - 0.1 × 9.81 = 0.981 m/s²
magnitude of angular acceleration = tangential acceleration / radius = 0.981 / 0.5 = 1.962 rad / s²
b) time for the train to come to rest = angular velocity / angular angular acceleration = 3.142/ 1.962 = 1.6 s
The equation earlier derived answer this question
Fnet + Ff = 0 since the body came to a rest
Fnet = - Ff and Ff = - μk mg, Fnet = ma
ma = - μk mg
m cancel m on both side
a = - μkg since it magnitude
a = μkg
