A wheel with rotational inertia i is mounted on a fixed, frictionless axle. the angular speed ω of the wheel is increased from zero to ωf in a time interval t. what is the average net torque τ on the wheel during this time interval?

Respuesta :

According to Newton's second law:
τ = Iα, where I is the moment of inertia and α is the angular acceleration.
Angular acceleration can be written as:
α = (ωf - ωi) / t
α = ωf/t

Substituting into the second law:
τ = Iωf/t

The average net torque τ on the wheel during time interval t is

[tex]{\tau= l( \dfrac{\omega_f}{t} )}[/tex].

Given to us:

Initial angular speed of the wheel, [tex]\bold{\omega_i} =0[/tex]

Final angular speed of the wheel, [tex]\bold{\omega_f} =\bold{\omega_f}[/tex]

time interval needed to increase the angular speed of the wheel, t = t

Angular acceleration of the wheel,

[tex]\alpha=\dfrac{\bold{\omega_f} -\bold{\omega_i}}{t},\\\alpha=\dfrac{\bold{\omega_f} -0}{t},\\\alpha=\dfrac{\bold{\omega_f}}{t}[/tex]

According to Newton's Second Law for Rotation:

If more than one torque acts on a rigid body about a fixed axis, then the sum of the torques equals the moment of inertia times the angular acceleration.

[tex]\begin{aligned}\\\sum_{0}^{\bold i}\tau&= \bold{l\alpha}\\\end{aligned}[/tex]

where,

τ is the torque

I is the moment of inertia,

α is the angular acceleration.

Putting the values into the second law, to calculate the average net torque τ on the wheel during this time interval;

[tex]\begin{aligned}\\\sum_{0}^{\bold i}\tau&= \bold{l\alpha}\\\tau&= \bold{l\alpha}\\&= l( \frac{\omega_f}{t} )\end{aligned}[/tex]

Hence, the average net torque τ on the wheel during time interval t is

[tex]{\tau= l( \dfrac{\omega_f}{t} )}[/tex].

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