A motorcycle accelerates uniformly from rest and reaches a linear speed of 24.8 m/s in a time of 9.87 s. The radius of each tire is 0.287 m. What is the magnitude of the angular acceleration of each tire

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:

=(V_f - v_i)/t

=(24.8-0)/9.87

=2.513 m/s²

Then;  the magnitude of angular acceleration

α = /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, , of the tires is equal to their linear acceleration, a. i.e

= a                  --------------(i)

But, the tangential acceleration, , is the product of the angular acceleration, , and the radius of the each of the tires. i.e

= r                  ------------(ii)

Combine equations (i) and (ii) as follows;

a = r                    --------------(iii)

Also, the linear acceleration, a, is given by;

a =                 ------------------(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;

= r          ------------------(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;

= 0.287

= 0.287

2.51 = 0.287

 =

= 8.75rad/s²

Therefore, the angular acceleration of each tire is 8.75rad/s²

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