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.
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Ответ:
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
ΔK = ½ 56.37 (52,569² - 111²)
ΔK = - 2.69 10⁵ J
Ответ:
ford f150
Explanation: