A disk is free to rotate around its fixed axis of revolution (delta). The moment of inertia of disk around delta is J=4×10^-2 kg.m^2
This disk is initially at rest is submitted to the action of a couple whose moment is equal to 0.2 N.m
What's the number of revolutions the disk performs during the first 2 seconds of motion
F = ma, and in rotational
Tau = I alpha
so alpha = .2/.04
then use eq of motion
theta = 1/2 alpha t^2
To determine the number of revolutions the disk performs during the first 2 seconds of motion, we need to calculate the angular acceleration of the disk first.
The torque (τ) applied to the disk is equal to the moment of inertia (J) multiplied by the angular acceleration (α):
τ = J * α
In this case, the torque applied to the disk is given as 0.2 N.m, and the moment of inertia is 4×10^-2 kg.m^2. We can rearrange the equation to solve for α:
α = τ / J
Plugging in the values, we get:
α = 0.2 N.m / 4×10^-2 kg.m^2
Now we need to calculate the angular velocity (ω) of the disk after 2 seconds of motion. The angular velocity is given by:
ω = α * t
Where t is the time. In this case, t is 2 seconds. Plugging in the values, we get:
ω = (0.2 N.m / 4×10^-2 kg.m^2) * 2 s
Now we can calculate the number of revolutions (N) based on the angular velocity. The number of revolutions is given by:
N = ω / (2π)
Where 2π represents a full revolution (360 degrees). Plugging in the values, we get:
N = [(0.2 N.m / 4×10^-2 kg.m^2) * 2 s] / (2π)
Simplifying the equation, we get:
N = (0.2 N.m * 2 s) / (4×10^-2 kg.m^2 * 2π)
Now we can calculate the final answer:
N = (0.4 N.m / 4×10^-2 kg.m^2π) = (0.1 / π) revolutions
So, the disk performs approximately (0.1 / π) revolutions during the first 2 seconds of motion.