5kg particle is atteched at a rim of arotating wheel of radius 2m. the wheel completes each revolution in 3 seconds. The total moment of inertia of the system(I partics+I wheel) is 2kg.m^2
A.determine the rotational kinetic energy of the system?
B.the work needed to stop the rotation?
C.how much torque is required to stop the rotation of the system in1.2second?
To answer the given questions, we need to understand some concepts about rotational dynamics.
The rotational kinetic energy (K_rot) of a rotating system is given by the formula:
K_rot = (1/2) * I * ω^2
where I is the moment of inertia and ω is the angular velocity.
a) To determine the rotational kinetic energy of the system, we need to find the angular velocity first. The wheel completes each revolution in 3 seconds, which means it has an angular velocity (ω) of 2π radians per 3 seconds, or (2π/3) radians per second.
We are given the total moment of inertia of the system (I_partics + I_wheel) as 2 kg.m^2.
Substituting the values into the formula, we have:
K_rot = (1/2) * 2 * (2π/3)^2
= (1/2) * 2 * (4π^2/9)
= 8π^2/18
= 4π^2/9 J
Therefore, the rotational kinetic energy of the system is 4π^2/9 Joules.
b) The work needed to stop the rotation of the system would be equal to the current rotational kinetic energy (K_rot) of the system.
The work-energy principle states that the work done on an object is equal to the change in its energy. In this case, the change in energy is from the rotating system to a stationary system.
So, the work needed to stop the rotation is equal to the rotational kinetic energy, which is 4π^2/9 J.
c) To calculate the torque required to stop the rotation of the system in 1.2 seconds, we can use the formula:
τ = Iα
where τ is the torque, I is the moment of inertia, and α is the angular acceleration.
The angular acceleration (α) can be calculated using the formula:
α = ωf - ωi / t
where ωf is the final angular velocity, ωi is the initial angular velocity, and t is the time taken.
Given:
ωi = (2π/3) rad/s (initial angular velocity)
ωf = 0 rad/s (final angular velocity)
t = 1.2 s (time taken)
Substituting the values, we have:
α = (0 - (2π/3)) / 1.2
= -(2π/3) / 1.2
= -5π/9 rad/s^2
Now, substituting the moment of inertia of the system (I = 2 kg.m^2) and the calculated angular acceleration (α = -5π/9 rad/s^2), we can calculate the torque (τ) using the formula:
τ = Iα
= 2 * (-(5π/9))
= -10π/9 N.m
Therefore, the torque required to stop the rotation of the system in 1.2 seconds is -10π/9 Newton meters.