Question

A car is fitted with an energy-conserving flywheel, which in operation is geared to the driveshaft so that the flywheel rotates at 240 rev/s when the car is traveling at 80 km/h. The total mass of the car is 950 kg; the flywheel weighs 200 N and is a uniform disk 1.1 m in diameter. The car descends a 1300-m-long, 5° slope, starting from rest, with the flywheel engaged and no power generated from the engine. Neglecting friction and the rotational inertia of the wheels, find

(a) the speed (m/s) of the car at the bottom of the slope,

(b) the angular acceleration (rad/s^2) of the flywheel at the
bottom of the slope, and

(c) the rate at which energy (W) is stored in the rotation of the
flywheel by the driveshaft as the car reaches the bottom of the
slope.

Answer #1

Initial gravitational potential energy= translation kinetic energy of car rotational + kinetic energy of flywheel

Mgh = 1/2(Mv^{2}) + (1/2) I
^{2}

I = (1/2) Mr^{2} for
a disk

ω = {(240 rev/s)(2π rad/rev) / [(80 km/h)(1000 m/km)(h / 3600 s)]}v

ω = 67.86 v

Mgh = (1/2)Mv² + (1/2)[(1/2)mr²](67.86v)²

MgLsinθ = (1/2)Mv² + (1151.2)mr²v²

(950 kg)(9.81 m/s²)(1300 m).sin5° = (1/2)(950 kg)v² + (1151.2)(20.4 kg)(.55 m)²v² [ Wheel weight 200 N = 200/9.8 KG =20.4 KG]

1.055e6 = 7579v² [ radius=1.1/2 =.55 m]

v = 11.79m/s

the speed of car is=11.79 m/s

..........................................................

the angular acceleration of the flywheel is ω ,

ω = {(240 rev/s)(2π rad/rev) / [(80 km/h)(1000 m/km)(h / 3600 s)]}v

ω = 67.86 v

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