Question

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

Answer 1

8.75rad/s²

Step-by-step explanation:

The tires of the motorcycle undergo a rolling motion. Therefore, the tangential acceleration,
`a_(T)`, of the tires is equal to their linear acceleration, a. i.e

`a_(T)` = a --------------(i)

But, the tangential acceleration,
`a_(T)`, is the product of the angular acceleration,
`\alpha`, and the radius of the each of the tires. i.e

`a_(T)` = r
`\alpha` ------------(ii)

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

a = r
`\alpha` --------------(iii)

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

a =
`(v - u)/(t)` ------------------(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;

`(v - u)/(t)` = r
`\alpha` ------------------(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;

`(24.8 - 0)/(9.87)` = 0.287
`\alpha`

`(24.8)/(9.87)` = 0.287
`\alpha`

2.51 = 0.287
`\alpha`

`\alpha` =
`(2.51)/(0.287)`

`\alpha` = 8.75rad/s²

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

Answer 2

8.756 rad/s²

Step-by-step 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:

`a_T` =(V_f - v_i)/t

=(24.8-0)/9.87

=2.513 m/s²

Then; the magnitude of angular acceleration

α =
`a_T` /r

=2.513/0.287

=8.756 rad/s²