A toy train rolls around a horizontal 1.0-{\rm m}-diameter track. The coefficient of rolling friction is 0.15. What is the magnitude of the train's angular acceleration after it is released?

Answer 1

150n

Answer 2

Well, Angular acceleration is just a measure of how many radians does the object's speed change by every second it moves around the center of a circle(α = rads/s^2). You are given a track radius, diameter, and coefficient of friction. F = Ma emulates motion in a straight line (or tangental motion). So the result of the following equation (F=mg*μ) is a force, and that for can be placed into 'F=Ma' to make a new equation: (mg*μ)=ma, but you've got to remember that all of this motion is in a straight line and we're being asked for angular (circular) motion. Now, we will substitute what we know into the equation: (m*9.81*0.15)=m*a....... and simplify: (m*1.47)=m*a.... then take the m out of both sides to equal: 1.47=acceleration (tangental). Now we have to convert tangential accel. to angular accel. by this equation: a = α*radius. Substituting it into the equation gives us this: 1.47 = α*0.5.... and α is what we're looking for, so we divide the radius out to equal:
α= 2.94 rad/s^2