A torque of 0.97Nm is applied to a bicycle wheel of radius 45cm and mass 0.60kg. Treating the wheel as a hoop, find its angular acceleration.

Question

anonymous · Answer

Have tried:$$T=1/2mr^{2}\alpha$$

anonymous · Answer

torque= moment of inertia * angular acceleration 
T=0.98Nm
r=45cm=0.45m
m=.60Kg
I=1/2mr^2=0.06075Kgm^2
a=T/I=16.13168ms^-2:)

anonymous · Answer

It needs to be in radians/s^{2}

anonymous · Answer

ya that i nt imp here:)

anonymous · Answer

That is not correct.

anonymous · Answer

it means the answer nt the unit:)

anonymous · Answer

I dont understand.

anonymous · Answer

Yo

anonymous · Answer

Your answer wasnt right. I did about the same thing and still couldnt get it.

anonymous · Answer

@halipearce  the answer is right............i garuntee:)

anonymous · Answer

No, I just  plugged it into mastering physics. It said it was wrong.

anonymous · Answer

$$\tau = I \alpha$$
I = mr^2

anonymous · Answer

Why is I=mr^2 when you are suppossed to treat the wheel like a hoop?

anonymous · Answer

yep

anonymous · Answer

moment for a hoop about an axis through its center is mr^2

anonymous · Answer

or through its center perpendicular to the plane of the hoop I should say...

anonymous · Answer

aka a wheel.