Easy rating for the person who can explain why ax alphaR hi

Easy rating for the person who can explain why a_x = -alpha*R (highlighted in blue). I just don\'t understand it, and the whole solution hinges on this.

Problem 1 The tire has a mass of 5 kg, and it rolls down an inclined surface. If it takes the tire 7 s to start from rest and toll 12 m down the surface, what is the tire\'s moment of inertia about its center of mass 330 mm 15° Solution: Free-Body Diagram ing 15 Equations of Motion F F mg sin b mar (i) 2 F N -mg cos 6 ma Kinematic Relationship a, 0;

Solution

solution;

1)here from given FBD diagram we have that for static equillibrium

Fx=max

-F+mgsin15=m*ax

for Fy=m*ay

N-mgcos15=m*ay

as in normal direction accelaration is zero hence resultant force is zero as it is virtual force which unable to perform work in real time.

so we get that

N=mgcos15=47.378 N

4)where for wheel equation for rotational motion is

-F*R+Ig*alpha=0

so we get that

F=Ig*alpha/R

5)on putting in that equation of Fx=m*ax we get that

-(Ig*alpha/R)+mgsin15=m*ax

6)where by equation of motion by newtos we get that

S=u*t+.5*ax*t^2

as u=0

we get

12=.5*ax*7^2

ax=.4897 m/s2

where for angular motion angular accelaration is given by

ax=alpha*R

where alpha=ax/r=.4897/.33=1.4839 rad/s2

note:linear accelaration=angular accealaration*Radius

[as x=arc length=radius*angle=R*theta

v=dx/dt=d(R.theta)/dt=R.d(theta)/dt

d(theta)/dt=w=angular velocity

v=R*w

ax=d(v)/dt=R.d(w)/dt=R*alpha

as alpha=dw/dt]

7)now initial equation is

-(Ig*alpha/R)+mgsin15=m*ax

on putting value we get moment of inertia of tire

Ig=2.2786 kg m2