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.
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.
To calculate the angular acceleration of the wheel, we can use the equation:
τ = Iα,
where τ is the torque applied to the wheel, I is the moment of inertia of the wheel, and α is the angular acceleration.
In this problem, the torque applied to the wheel is equal to the product of the applied force and the radius of the wheel, i.e., τ = Fr. In this case, the force applied is 50.0 N and the radius of the wheel is 0.400 m.
Thus, τ = (50.0 N)(0.400 m) = 20.0 N·m.
Now, we can substitute this value for torque into the equation:
20.0 N·m = (4.00 kg·m^2)α.
To solve for α, we rearrange the equation:
α = (20.0 N·m) / (4.00 kg·m^2).
Now let's calculate it:
α = 5.0 N·m / kg·m^2.
Therefore, the angular acceleration of the wheel is 5.0 N·m / kg·m^2.
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