Answer:
Step-by-step explanation:
We can use the work-energy principle to find the work done by the applied force to stop the disk. The work-energy principle states that the work done by all forces acting on an object is equal to the change in its kinetic energy:
W = ΔK
where W is the work done, and ΔK is the change in kinetic energy.
Initially, the disk is rotating with an angular velocity of 950 rev/min. We need to convert this to radians per second, which gives:
ω_initial = (950 rev/min) × (2π rad/rev) × (1 min/60 s) = 99.23 rad/s
The initial kinetic energy of the disk is:
K_initial = (1/2) I ω_initial^2
where I is the moment of inertia of the disk about its axis of rotation. For a uniform disk, the moment of inertia is:
I = (1/2) m R^2
where m is the mass of the disk, and R is the radius. Substituting the given values, we get:
I = (1/2) (190 kg) (1.1 m)^2 = 115.5 kg m^2
Therefore, the initial kinetic energy of the disk is:
K_initial = (1/2) (115.5 kg m^2) (99.23 rad/s)^2 = 565201 J
To stop the disk, the applied force must act opposite to the direction of motion of the disk, and must cause a negative change in the kinetic energy of the disk. The force is applied at a radial distance of 0.5 m, which gives a torque of:
τ = F r
where F is the magnitude of the force. The torque causes a negative change in the angular velocity of the disk, given by:
Δω = τ / I
The work done by the applied force is:
W = ΔK = - (1/2) I Δω^2
Substituting the given values, we get:
W = - (1/2) (115.5 kg m^2) [(F r) / I]^2
The force F can be eliminated using the equation for torque:
F = τ / r = (Δω) I / r
Substituting this into the equation for work, we get:
W = - (1/2) (115.5 kg m^2) [(Δω) I / r I]^2
= - (1/2) (115.5 kg m^2) (Δω / r)^2
Substituting the values for Δω and r, we get:
W = - (1/2) (115.5 kg m^2) [(F r / I) / r]^2
= - (1/2) (115.5 kg m^2) [(2 Δω / R) / (2/5 m R^2)]^2
= - (1/2) (115.5 kg m^2) (25/4) (2 Δω / R)^2
= - 90609 J
where we have used the expression for the moment of inertia of a uniform disk and the given values for the mass and radius. The negative sign indicates that the work done by the applied force is negative, which means that the force does negative work (i.e., it takes energy away from the system). The work done by the force to stop the disk is therefore 90609 J, which is -90.6 kJ (to two decimal places).