Respuesta :
Answer:
a) I= 0.0183 kg.m²
b) τ = 3.846 × 10⁻³ N.m
c) [tex]a_c[/tex] = 0.07315 m/s²
d) 62 revolutions
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
[tex]\alpha = \frac{\omega - \omega_o}{t}[/tex]
on substituting the values, we have
[tex]\alpha = \frac{0 - 12.56}{60}[/tex]
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:
[tex]a_c=r\alpha[/tex]
on substituting the values, we get
[tex]a_c=0.35\times0.209[/tex] = 0.07315 m/s²
d) To calculate the revolutions made by the wheel while coming to rest is given as:
[tex]\theta=\omega_ot-\frac{1}{2}\alpha t^2[/tex]
on substituting the values, we get
[tex]\theta=12.56\times 60-\frac{1}{2}12.56 \times60^2[/tex]
or
[tex]\theta=377.4[/tex]rad
or
[tex]\theta=377.4\times\frac{1}{2\pi}[/tex]revolutions
or
60 revolutions
thus, total revolutions made is 60 + 2 = 62 revolutions.
