Asked by Sunny
A string is wrapped around the rim of a flywheel 0.400 m in radius, and a steady pull of 50.0 N is exerted on the cord. The wheel is mounted in frictionless bearings on a horizontal shaft through it's center. The moment of inertia of the wheel is 4.00 kg.m^2 .Calculate the angular acceleration of the wheel.
Answers
Answered by
MathMate
Read up examples in
http://spiff.rit.edu/classes/phys211/lectures/rotke/rotke_all.html
one of which is similar to this one.
After this, you can attempt the previous problem you posted, together with your previous knowledge of kinematics and energy principles.
http://spiff.rit.edu/classes/phys211/lectures/rotke/rotke_all.html
one of which is similar to this one.
After this, you can attempt the previous problem you posted, together with your previous knowledge of kinematics and energy principles.
Answered by
Benson Yope
The torque T = 50N*0.4m
= 20N.m
The moment of inertia is I = 4.00kg.m^2.
The angular acceleration in rad/s^2 alpha = T/I
=20N.m/(4.00kg/m^2)
=5 rad/s^2
= 20N.m
The moment of inertia is I = 4.00kg.m^2.
The angular acceleration in rad/s^2 alpha = T/I
=20N.m/(4.00kg/m^2)
=5 rad/s^2
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