Question

The combination of an applied force and a frictional force produces a constant torque of 40 N·m on a wheel rotating about a fixed axis. The applied force acts for 8.0 seconds, during which time the angular speed of the wheel increases from 0 to 700 degrees/second. The applied force is then removed, and the wheel comes to rest in 55 s. Answer the following questions. (a)What is the magnitude of the angular acceleration of the wheel while the applied force is acting on the wheel? 87.7 Correct: Your answer is correct. degrees/second2 (b) What is the moment of inertia (rotational inertia) of the wheel? 26.1 Incorrect: Your answer is incorrect. kg · m2 (c)What is the magnitude of the angular acceleration of the wheel while only the frictional force is acting on the wheel? degrees/second2 (d) What is the magnitude of the frictional torque? N·m (e) How many revolutions does the wheel make while the applied force is acting on the wheel? revolutions (f) How many revolutions does the wheel make while only friction is acting on it? revolutions (g) How many revolutions does the wheel?make during this whole proces

Answer 1

a.

Angular acceleration = 87.7 degree/s2 = 87.7×2π/180 = 3.06 rad/s2

b.

As we know

Torque = momentum of inertia × angular acceleration

40 = I×3.06

I = 13.07 kgm2​​​​​​

C.

When oly frictional force

By equation of motion

0 = 700 - α×55

Angular acceleration (α) = 12.72 degrees/s2

= 12.72×2π/180 = 0.44 rad/s2​​​​​

d.

Frictional torque = Iα = 13.07×0.44 = 5.8 Nm

e.

By equation of motion

(700)2 = 0 + 2×87.7×angle

Angle = 2793.6 degree

In revolutions = 2793.6/360 =7.76 revolutions

f.

By equation of motion

0 = (700)2 - 2×12.72×angle

Angle = 19261 degrees

360 degrees = 1 revolution

19261 degrees = 19261/360 = 53.5 revolutions

g.

Total revolution = 7.76 + 53.5 = 61.26 revolutions