Question

You want to figure out the friction that acts on the rear wheel of your bike. You give the wheel a quick push and find that it spins 2 complete revolutions in the first second and takes 60.0 seconds to come to rest. (a) If the wheel has a mass 0.15 kg and a radius of 0.35 m, what is the moment of inertia, assuming all the mass is located at the radius? (b) If the frictional torque is constant, estimate the value of this torque? (c) What is theradial acceleration of a point on the tire? (d) How many revolutions did the wheel make before coming to rest??

Answer 1

a) I= 0.0183 kg.m²

b) τ = 3.846 × 10⁻³ N.m

c)
`a_c` = 0.07315 m/s²

d) 62 revolutions

Step-by-step explanation:

Given:

Mass of the wheel, m = 0.15 kg

radius of the wheel. r = 0.35 m

Angular speed, ω = 2 rev/s = 2 × 2π rad/s = 4π rad/s

a) The moment of inertia for a wheel is given as:

I = mr² = 0.15 × 0.35²

or

I= 0.0183 kg.m²

b) The torque is given as:

τ = Iα

where α is the angular acceleration

`\alpha = (\omega - \omega_o)/(t)`

on substituting the values, we have

`\alpha = (0 - 12.56)/(60)`

or

α = -0.209 rad/s²

substituting the α in the formula for torque, we get

τ = 0.15 × 0.35² × 0.209 = 3.846 × 10⁻³ N.m

c) The centripetal acceleration is given as:

`a_c=r\alpha`

on substituting the values, we get

`a_c=0.35*0.209` = 0.07315 m/s²

d) To calculate the revolutions made by the wheel while coming to rest is given as:

`\theta=\omega_ot-(1)/(2)\alpha t^2`

on substituting the values, we get

`\theta=12.56* 60-(1)/(2)12.56 *60^2`

or

`\theta=377.4`rad

or

`\theta=377.4*(1)/(2\pi)`revolutions

or

60 revolutions

thus, total revolutions made is 60 + 2 = 62 revolutions.