Aflywheel in the form of a heavy circular disk of diameter 0.612 m and mass 301 kg is mounted on a frictionless bearing. a motor connected to the flywheel accelerates it from rest to 1060 rev/min. a) what is the moment of inertia of the flywheel? answer in units of kg · m2 .b)how much work is done on it during this acceleration? answer in units of j. c)after 1060 rev/min is achieved, the motor is disengaged. a friction brake is used to slow the rotational rate to 502 rev/min. what is the magnitude of the energy dissipated as heat from the friction brake? answer in units of j.

a)   I= 56.37 kg m², b)  W = 3.47 10⁵ J

c) ΔK = - 2.69 10⁵ J

Explanation:

a) The moment of inertia of the steering wheel is the moment of inertia of a disc with a shaft that passes through its center

      I = ½ M R²

      I = ½ 301  0.612²

      I= 56.37 kg m²

b) Work is equal to the change in kinetic energy

     W = ΔK = ½ I w² - ½ I w₀²

As part of the rest the initial angular velocity is zero

     W = ½ I w²

Let's reduce the angular velocity to SI units

     w = 1060 rev / min (2π rad / 1rev) (1min / 60s) = 111 rad / s

     W = ½ 56.37 111²

     W = 3.47 10⁵ J

c) Dissipated energy is equal to work

    W = ΔK

    W = ½ I ² - ½ I w₀²

For this case

    w₀ = 111 rad / s

      = 502 rav / min (2pirad / 1rev) (1min / 60s) = 52,569 rad / s

     ΔK = ½ 56.37 (52,569² - 111²)

     ΔK = - 2.69 10⁵ J

answer: what are you talking about?

explanation:

inverse v=i*r

nice nice