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.
Torque = force applied * lever arm
= 50.0 N * 0.400m
= _____ N*m
Torque = angular acceleration * moment of inertia
You know the torque and are given the moment of inertia; substitute these into the above equation and solve for the angular acceleration.
To calculate the angular acceleration of the wheel, you can use Newton's second law of motion for rotating objects, which states that the net torque acting on an object is equal to the product of its moment of inertia and its angular acceleration.
The formula for calculating the angular acceleration (α) is:
α = τ / I
where:
α is the angular acceleration,
τ is the net torque acting on the object,
I is the moment of inertia of the wheel.
In this case, the moment of inertia (I) of the wheel is given as 4.00 kg.m^2.
To calculate the torque (τ), you can use the equation:
τ = F * r
where:
τ is the torque,
F is the force applied,
r is the distance from the point where the force is applied to the axis of rotation.
In this case, the force (F) applied is 50.0 N, and the radius (r) of the wheel is 0.400 m.
Substituting these values into the equation, we have:
τ = 50.0 N * 0.400 m = 20.0 N.m
Now, we can calculate the angular acceleration (α) using the formula:
α = τ / I = 20.0 N.m / 4.00 kg.m^2 = 5.00 rad/s^2
Therefore, the angular acceleration of the wheel is 5.00 rad/s^2.