Answer:
8.756 rad/s²
Explanation:
Given that:
A motorcycle accelerates uniformly from rest, then initial velocity v_i = 0 m/s
It final velocity v_f = 24.8 m/s
time (t) = 9.87 s
radius (r) of each tire = 0.287 m
Firstly; the linear acceleration of the motor cycle is determined as follows:
[tex]a_T[/tex] =(V_f - v_i)/t
=(24.8-0)/9.87
=2.513 m/s²
Then; the magnitude of angular acceleration
α =[tex]a_T[/tex] /r
=2.513/0.287
=8.756 rad/s²
8.75rad/s²
The tires of the motorcycle undergo a rolling motion. Therefore, the tangential acceleration, [tex]a_{T}[/tex], of the tires is equal to their linear acceleration, a. i.e
[tex]a_{T}[/tex] = a --------------(i)
But, the tangential acceleration, [tex]a_{T}[/tex], is the product of the angular acceleration, [tex]\alpha[/tex], and the radius of the each of the tires. i.e
[tex]a_{T}[/tex] = r[tex]\alpha[/tex] ------------(ii)
Combine equations (i) and (ii) as follows;
a = r[tex]\alpha[/tex] --------------(iii)
Also, the linear acceleration, a, is given by;
a = [tex]\frac{v - u}{t}[/tex] ------------------(iv)
Where;
v = final linear speed of the tire
u = initial linear speed of the tire
t = time taken for the motion
Combine equations iii and iv as follows;
[tex]\frac{v - u}{t}[/tex] = r[tex]\alpha[/tex] ------------------(v)
From the question;
v = 24.8m/s
u = 0 (since the motorcycle accelerates from rest)
t = 9.87s
r = 0.287m
Substitute these values into equation (v) as follows;
[tex]\frac{24.8 - 0}{9.87}[/tex] = 0.287[tex]\alpha[/tex]
[tex]\frac{24.8}{9.87}[/tex] = 0.287[tex]\alpha[/tex]
2.51 = 0.287[tex]\alpha[/tex]
[tex]\alpha[/tex] = [tex]\frac{2.51}{0.287}[/tex]
[tex]\alpha[/tex] = 8.75rad/s²
Therefore, the angular acceleration of each tire is 8.75rad/s²
