Torque generated by the motor of ferris wheel?

[jeneekim]

Homework Statement

An electric motor can accelerate a Ferris
wheel of moment of inertia 11300 kg · m2 from
rest to 9.6 rev/min in 14.6 s. When the mo-
tor is turned off, friction causes the wheel to
slow down from 9.6 rev/min to 8.37 rev/min
in 7.23 s.
Determine the torque generated by the mo-
tor to bring the wheel to 9.6 rev/min.
Answer in units of N · m.

Homework Equations

$$\alpha$$ = $$\omega$$/t
$$\tau$$ = I$$\alpha$$

The Attempt at a Solution

given:
I = 11300 kgm²
$$\omega$$_o = 0
$$\omega$$_f = 9.6 rev/min = 1.005309649 rad/s
t_f = 14.6 s

The question asks to determine the torque generated by the motor to bring the wheel to 9.6 rev/min so I think that means I don't have to worry about the motor beign turned off or the friction to slow it down... sooooo...

$$\alpha$$ = $$\omega$$/t
= (1.005309649 rad/s)/(14.6s)
= 0.0688568253 rad/s²

$$\tau$$ = I$$\alpha$$
= (11300kgm²)(0.0688568253 rad/s²)
= 778.0821257 Nm

When I enter this answer, it says it is wrong? What am I doing wrong here?

Thank you in advance for your help!

 

[rl.bhat]

In the first case
T = I*αo + I*αr
where I*αr is the torque required to overcome the friction and I*αo is the torque need to accelerate the wheel after overcoming the frictional force.
In the second case
Tf = I*αr.

 

[kerimek]

Your approach and calculations are correct. However, the answer should be given in units of N*m, not Nm. The correct answer is 778.0821257 N*m.