1) The combination of an applied force and a frictional force produces a constan

Question

1) The combination of an applied force and a frictional force produces a constant torque of 30 N · m on a wheel rotating about a fixed axis. The applied force acts for 7.5 s, during which time the angular speed of the wheel increases from 0 to 17 rad/s. What is the moment of inertia of the wheel? Answer in units of kg · m^2

2) The applied force is then removed, and the wheel comes to rest in 70 s. What is the frictional torque? Answer in units of N · m.

3) How many revolutions does the wheel make during the entire 78 s interval? Answer in units of rev.

Explanation / Answer

Torque=I*alpha

also

w=wi+alpha*t

given that w=17

t=7.5

wi=0

from the problem statement

17=alpha*7.5

alpha=17/7.5 =2.26

put into torque equation

30=I*2.26

I=30/2.26

I=13.27

b) similar, only a new torque and a new alpha

0=17+alpha*70

now w = 0 .wi=17, t=70s

alpha = -17/70 = -0.24

put it again in torque equation

T=13.70*(-0.24)

T=-3.32 Nm

c)

theta=theta0+wi*t+.5*alpha*t^2

theta is the angle of displacement

th=(0.5*17*(7.5^2))

for the first 7.5 seconds

w = (0.5*17*(7.5^2))/7.5

notice how this is the average w times t

w = 0.5*17*7.5

w=63.75 rad

the next 70 s for the total

theta = (63.75+17*70-0.5*17*70^2)/70

or

theta = 63.75 + 17*70 - 0.5*17*70

theta = 658.75 rad

there are 2*pi rad per revolution
so revolutions = 658.75 / (2*3.14)

104.89 revolutions.

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