The combination of an applied force and a frictional force produces a constant t

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 8.7 s, during
which time the angular speed of the wheel
increases from 0 to 16 rad/s.
a)What is the moment of inertia of the wheel?
Answer in units of kg · m2

b)The applied force is then removed, and the
wheel comes to rest in 56 s.
What is the frictional torque?
Answer in units of N · m

c)How many revolutions does the wheel make
during the entire 65 s interval?
Answer in units of rev

Explanation / Answer

(a)

Angular acceleration = 16-0 / 8.7 = 1.84 rad/s2

Torque = Moment of Inertia x Angular acceleration

implies Moment of Inertia = 40 / 1.84 = 21.74 kg.m2

(b)

final = initial + t

0 = 16 + (56)

= -0.286 rad/s2

So Frictional torque = Moment of inertia x 0.286 = 21.74 x 0.286 = 6.218 N.m

(c)

final2 - initial2 = 2

0 - 162 = 2 x -0.286 x

= 256/(2x0.286) = 447.55 radians = 447.55/2 revolutions = 71.23 revolutions

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---