There's a question that asks: if a wheel with 1.56 m/rad radius which reached a final velocity of 200 rad/min(or 10/3 rad/s) were slowed down with a constant torque of 10,000 NM being applied, how long would it take to stop the wheel. I can't seem to find how torque is related to this...
Possibility for Part C
Angular Displacement = (W^2-Wo^2)/(2*Angular Acceleration)
= (11.11rad^2/s^2)/(4.44rad/s^2)
= 2.5 rad
Angular Displacement = (t/2)(Wo+W)
=(2.5 rad) = (t/2)*(10/3rad/s)
=(2.5 rad)/(3.33rad/s)
= .75s = (t/2)
t = 1.5 s
Now, I think I solved that part, but I didn't...
Ok, here's what I'm thinking.
a.) (200/60)rad/min = (10/3)rad/s
Angular Acceleration = (dW/dT)
(3.33rad/s)/(1.5s) = 2.22 rad/s^2
b.) I = MR^2 F = Ma
5000N = (M)(2.22rad/s^2)
M = 2252.25 kgm/rad
I = MR^2
(5500kgm^2) = (2252.25kgm/rad)(R^2)
R = 1.56 m/rad
c.) No idea. Please Help...
Heres the problem. I'm trying to solve it at the moment, and I'll be posting what I have done so far. I'm very short on formulas for this so I will need help. Thanks a lot.
1) a) Determine angular acceleration of a wheel, which has a velocity of 200 rad/min after rotating for 1.5s from...
Maybe for part C I can do this...
Angular Displacement = (W^2-Wo^2)/(2*Angular Acceleration)
= (11.11rad^2/s^2)/(4.44rad/s^2)
= 2.5 rad
Angular Displacement = (t/2)(Wo+W)
=(2.5 rad) = (t/2)*(10/3rad/s)...
Ok, here's what I'm thinking.
a.) (200/60)rad/min = (10/3)rad/s
Angular Acceleration = (dW/dT)
(3.33rad/s)/(1.5s) = 2.22 rad/s^2
b.) I = MR^2 F = Ma
5000N = (M)(2.22rad/s^2)
M = 2252.25 kgm/rad
I = MR^2
(5500kgm^2) = (2252.25kgm/rad)(R^2)
R = 1.56...
I missed a week of school because of a virus. During that week, I missed taking a lot of physics notes. When I returned, I received a problem from my teacher that he wants me to solve. I have the notes now but I don't understand them because they weren't taught to me, so may someone please help...