Question

A toy train rolls around a horizontal 1.0-m-diameter track.The coefficient of rolling friction is 0.10.a) What is the magnitude of the trains angular acceleration afterit is released?b)How long does it take the train to stop if it's released with anangular speed of 30 rpm?In step one of the question, the solution supplied states thattangential acceleration is the coefficient of rolling frictionmultiplied by gravity. Why is this so?

Answer 1

1.962 rad / s², 1.6 s

Step-by-step 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