Question

Consider an old-fashioned bicycle with a small wheel of radius 0.12 m and a large wheel of radius 0.8 m. Suppose the rider starts at rest, accelerates with a constant acceleration for 1 minutes to a velocity of magnitude 15.6 m/s. He maintains his velocity for 10.9 minutes and then decelerates, with constant deceleration, for 4.1 minutes. Find the total angle through which a point on the small wheel has turned.

Answer 1

we have to calculate this problem, first distance covered. _To_ solve the total I distance covered in constant acceleration for the V = u= to 15.6 mls 0 mis 1 min = 60 seconds - - ve ut at_ a = 15.6 60 0 m/s² = 0.26 m/s². S : ut + at -30+ x0.25 x 602) = 468 m dis tance covered motion with constant velocity. V: 15.6 ms _time ,t e 10.9 min. -- 10.9 x60 s __.= 654 s distance covered S2 = vxt = 15:6 x 654 10202.4._ m m

IL _distance covered during _deceleration. us_initial velocity - = 15.6 m/s ve final velocity =0 m/s t = time taken = 4.1 min. - 4.1 x60 s - 246 S av = ut at => 0 = 15.6 + (ax 246) > a = -15-6____ m/s2_ 246 = -0.06 3415_m/s.? sz= ut + 1 ot? = (15.6 x 246) +11x -0.063 415 x 2462 = 191.8.8 m Total bicycle distance covered by = Sit S2 + S3 = 468 + 10202.4. S 12589-2 m t 1918.8 m

Circumference of small wheel = 278 = 271 X 0.12 __m = 0.7536 m No. of revolutions made by small wheel Total distance covered Curcumference 12589.2 revolutions 0.7536 16705.414 revolutions __ Total has turned angle through which small wheel = Ne of_ revolutions x 20 = 16705. 414_x 2x 3.14 104 910 radians (approx.) stadians
If you have doubt in any part of the solution, feel free to ask.

Answer 2

SOLUTION :

v = 15.6 m/s in 1 min (60sec) 

Acceleration, a = ∆v / t = (15-6 - 0) / 60 = 0.26 m/s^2

Distance covered during constant acceleration , s1 

= average velocity * time

= (15.6 - 0)/2 * 60

= 468 m 

Distance covered at constant speed for 10.9 min, s2

= speed * time

= 15.6 * 10.9 * 60

= 10202.40 m

Time of deceleration = 4.1 min (246 sec)

Average speed during deceleration = 15.6 / 2 = 7.8 m/s

Distance covered during deceleration, s3

= average speed * time

= 7.8 * 246

= 1918.80 m 

Total distance covered, s

= s1 + s2 + s3

= 468 + 10202.40 + 1918.80 

= 12589.20 m 

Number of revolutions made by small wheel, N = s / (pi d) 

Small wheel turns angle through for this 

= N * 2 pi 

= s/(pi d) * 2pi 

= 2s/d

= 2*12589.20 / (0.12*2)

= 104910  rad. (ANSWER)