# A bicycle wheel of radius r= 0.25 m starts at rest and rolls 150 m without slipping in 30 s. Calculate a) the number of revolutions the wheel makes, b) the number of radians through which it turns, c) the average angular velocity.Assuming the angular acceleration of the wheel given above was constant, at the end of one revolution calculate: d) the angular acceleration, e) the final angular velocity, and f) the tangential velocity and the tangential acceleration of a point on the rim.

a)

Distance for one revolution  = 2*pi*r

= 2*pi*0.25

=pi/2

Number of revolution to travel 150m = 150/(pi/2)

= 95.49 rev

b)

Therefore the number of radians it turns = 2*pi*95.49

= 600

c)

average angular velocity =...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

a)

Distance for one revolution  = 2*pi*r

= 2*pi*0.25

=pi/2

Number of revolution to travel 150m = 150/(pi/2)

= 95.49 rev

b)

Therefore the number of radians it turns = 2*pi*95.49

= 600

c)

average angular velocity = (total radians turned)/time taken

= 600/30

d)

From equations for angular motion;

` `` ``theta = omega*t+0.5*alpha*t^2`

Since the bicycle starts from rest ` ``omega` = 0

at time t=30 theta = 600 rad

Therefore;

` 600 = 0*30+0.5*alpha*30^2`

`alpha = 1.33333`

e)

From equations for angular motion;

`omega = omega_0+alpha*t`

when t=30;

So angular velocity at end=40 rad/s

f)

Tangential velocity (v)       = `r*omega`

tangential accalaration (a) = `r*omega^2`

consider final velocity;

v = 0.25*40 = 10m/s

a = 0.25*40^2 = 400m/s

Approved by eNotes Editorial Team