A 16 kg mass is attached to a cord that is wrapped around a wheel with a radius of 10.9cm. The acceleration of the mass down the frictionless incline is measured to be 1.4 m/s2.The angle of incline is 36 degrees. Assume the axle of the wheel to be frictionless. The force of the the rope in 69.77N. the inertia is .5923.

Find the angular speed of the wheel 2.0 s after it begins rotating, starting from rest. Answer in units of rad/s.

What are the units of what you call "the inertia"? Is that supposed to be the MOMENT OF inertia of the wheel? Do you know what that means?

If you know the acceleration rate of the mass going dwn the incline, you should be able to say how fast the wheel is turning after 2.0 seconds, without most of the other information.

To find the angular speed of the wheel 2.0 seconds after it begins rotating, we need to use the principle of conservation of energy. The total mechanical energy of the system remains constant, so the initial potential energy of the mass on the incline is converted into rotational kinetic energy of the wheel.

The potential energy of the mass on the incline is given by:
Ep = m * g * h

where m is the mass (16 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the incline. We can calculate h using the angle of incline (θ) and the radius of the wheel (r):

h = r * sin(θ)

Substituting the values, we have:
h = 0.109 m * sin(36°) ≈ 0.0658 m

The potential energy (Ep) can then be calculated as:
Ep = 16 kg * 9.8 m/s^2 * 0.0658 m ≈ 10.144 J

This potential energy is converted into rotational kinetic energy (Er) of the wheel. The rotational kinetic energy is given by:
Er = (1/2) * I * ω^2

where I is the moment of inertia of the wheel and ω is the angular speed. We are given the moment of inertia (I = 0.5923) and we need to find ω.

We know that the force applied by the rope (F) is equal to the torque (τ) acting on the wheel, which is given by:
τ = F * r

Substituting the values, we have:
τ = 69.77 N * 0.109 m ≈ 7.608 J

The torque is also given by:
τ = I * α

where α is the angular acceleration. Rearranging the equation, we have:
α = τ / I

α is also related to the angular speed (ω) and time (t) by the equation:
ω = α * t

Rearranging again, we have:
α = ω / t

Substituting the known values, we can solve for ω:
ω = (τ / I) * t = (7.608 J / 0.5923) * 2.0 s ≈ 20.42 rad/s

Therefore, the angular speed of the wheel 2.0 seconds after it begins rotating is approximately 20.42 rad/s.