Physics: Moment of inertia? Help today!!?

Question

The combination of an applied force and a
frictional force produces a constant torque of
25 Nm on a wheel rotating about a fixed
axis. The applied force acts for 7.8 s, during
which time the angular speed of the wheel
increases from 0 to 14 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 77 s.
What is the frictional torque? Answer in Nm.
c)How many revolutions does the wheel make
during the entire 85 s interval? Answer in
units of rev.

O__o;; I have a midterm soon, and I'd appreciate it if someone could get back to me soon so I can study how to do this right! I don't know if what I have is even the right thing to start with. thanks!

Answer 1

Torque=I*alpha
also
w=wi+alpha*t

from the problem statement
14=alpha*7.8
alpha=14/7.8

plug into torque equation
25=I*14/7.8
I=25*7.8/14
I=13.93

b) similar, only a new torque and a new alpha
0=14+alpha*77
alpha=-14/77
T=-13.93*14/77
T=-2.53 Nm

c)
th=th0+wi*t+.5*alpha*t^2
th is the angle of displacement

for the first 7.8 seconds
th=.5*14*7.8^2/7.8
notice how this is the average w times t
.5*14*7.8

54.6 rad
the next 77 s for the total
th=54.6+14*77-.5*14*77^2/77
or
th=54.6+14*77-.5*14*77
th=593.6 rad

there are 2*pi rad per revolution
so revolutions=593.6/(2*3.14)

94.5 revolutions

j

Answer 2

hi Diamond! Does it advise you're doing a lab? if so then the linked fee of .0195 is plenty greater possibly to be closer to real one from above, because of the fact the kinetic friction is often below static friction. If it is real then your “fairly vast” pulley might have the same share of reliability of your effect, on the same time as actual the distinction in instruments of kg*m^2 may be greater than that for a lighter pulley.