Question

A bicycle racer is going downhill at 10.4 m/s when, to his horror, one of his 2.34 kg wheels comes off when he is 68.0 m above the foot of the hill. We can model the wheel as a thin-walled cylinder 85.0 cm in diameter and neglect the small mass of the spokes.How fast is the wheel moving when it reaches the foot of the hill if it rolled without slipping all the way down?

Answer 1

Answer:

$v_1 = 27.83 m/s$

Explanation:

Given data:

downhill velocity = 10.4 m/s

weight of wheel is 2.34 kg

distance from foot of hill is 68 m

cylinder diameter is 85 cm

from conservation of energy principle we have following relation

$mgh + \frac{1}{2} mv^2 +\frac{1}{2} I_{cm} \omega^2 =

mgh + \frac{1}{2} mv_1^2 +\frac{1}{2} I_{cm} \omega^2$

we know that moment of inertia $I_{cm} = mR^2$

                                                                  $= 2.34\times (\frac{0.85}{2})$

                                                                 = 0.99  m^4

$\omega = \frac{v}{R}$

SO WE HAVE

$mgh + \frac{1}{2} mv^2 +\frac{1}{2} I_{cm} \frac{v}{R}^2 =

mgh + \frac{1}{2} mv_1^2 +\frac{1}{2} I_{cm} \frac{v_1}{R}$

$mgh + mv^2 = mv_1^2

[tex]v_1^2 = v^2 + gh$

$= 10.4^2 + 9.8\times 68$

$v_1 = 27.83 m/s$