Answer:
the moment of inertia of the wheel is 0.408 kg.m²
Explanation:
Given;
tangential force applied to the wheel, f = 90 N
radius of the wheel, r = 0.15 m
initial angular speed of the wheel, ω₁ = 0
final angular speed of the wheel, ω₂ = 14.3 rev/s
time of motion of the wheel, t = 2.72 s
The tangential acceleration of the wheel is calculated as;
[tex]\alpha _t = \alpha r[/tex]
where;
[tex]\alpha[/tex] is the angular acceleration
[tex]\alpha = \frac{\Delta \omega }{\Delta t} = \frac{\omega _2 - \omega_1}{t_2-t_1}\\\\where;\\\\\omega_2 \ is \ the \ final \ angular \ speed = 14.3 \frac{rev}{s} \times \frac{2\pi \ rad}{1 \ rev} = 89.86 \ rad/s[/tex]
[tex]\alpha = \frac{89.86 - 0}{2.72} = 33.04 \ rad/s^2\\\\\alpha _t = \alpha r\\\\\alpha _t = 33.04 \ rad/s^2 \times 0.15 \ m\\\\\alpha _t = 4.96 \ m/s^2[/tex]
The mass of the wheel is calculated as;
F = ma
m = F/a
m = (90)/(4.96)
m = 18.15 kg
The moment of inertia of the wheel is calculated as;
I = mr²
I = 18.15 x (0.15)²
I = 0.408 kg.m²
Therefore, the moment of inertia of the wheel is 0.408 kg.m²
The moment of inertia of the rotating wheel without friction will be I=0.408 kg.m²
The Moment of inertia, denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis,
It is given in the question
The tangential force applied to the wheel, f = 90 N
The radius of the wheel, r = 0.15 m
The initial angular speed of the wheel, ω₁ = 0
The final angular speed of the wheel, ω₂ = 14.3 rev/s
time of motion of the wheel, t = 2.72 s
The tangential acceleration of the wheel is calculated as;
[tex]\alpha _t=\alpha r[/tex]
[tex]\alpha[/tex] is the angular acceleration
[tex]\alpha=\dfrac{(w_2-w_1)}{(t_2-t_1)}[/tex]
[tex]w_2=14.3\times 2\pi=89.8\ \dfrac{rad}{s}[/tex]
[tex]\alpha=\dfrac{89.86-0}{2.72} =33.01\ \dfrac{rad}{s}[/tex]
The mass of the wheel is calculated as;
[tex]F=ma[/tex]
[tex]m =\dfrac{F}{a}[/tex]
[tex]m=\dfrac{90}{4.96}[/tex]
[tex]m=18.15 kg[/tex]
The moment of inertia of the wheel is calculated as;
[tex]I=mr^2[/tex]
[tex]I=18.15\times (0.15)^2=0.408\ kgm^2[/tex]
Thus the moment of inertia of the rotating wheel without friction will be I=0.408 kg.m²
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