(a) To find the time it takes to bring the DVD to a stop when a friction force of 0.5 N is applied at the outer edge of the disk, we can use the equation for rotational motion:
τ = I * α
Where:
τ is the torque applied to the disk,
I is the moment of inertia of the disk,
α is the angular acceleration of the disk.
The moment of inertia of a solid disk is given by:
I = (1/2) * m * r^2
Where:
m is the mass of the disk,
r is the radius of the disk.
Substituting the given values:
m = 16 g = 0.016 kg
r = 6 cm = 0.06 m
I = (1/2) * 0.016 kg * (0.06 m)^2
I ≈ 3.6 × 10^(-5) kg·m²
The torque applied to the disk is equal to the friction force multiplied by the radius:
τ = F * r
τ = 0.5 N * 0.06 m
τ = 0.03 N·m
Using the equation τ = I * α, we can solve for α:
α = τ / I
α = 0.03 N·m / 3.6 × 10^(-5) kg·m²
α ≈ 833.3 rad/s²
Now, we can use the equation for angular acceleration:
α = Δω / Δt
Where:
Δω is the change in angular velocity,
Δt is the time taken.
Since the DVD starts at 1000 revolutions per minute (rpm) and comes to a stop, Δω is equal to the final angular velocity (0) minus the initial angular velocity:
Δω = 0 - (1000 rpm * 2π rad/rev * 1 min/60 s)
Converting units:
Δω = -1047.2 rad/s
Now we can solve for Δt:
Δt = Δω / α
Δt = (-1047.2 rad/s) / 833.3 rad/s²
Δt ≈ -1.26 s
The time it takes to bring the DVD to a stop is approximately 1.26 seconds.
(b) If the friction force is applied just 1 cm away from the center of the disk, the moment of inertia changes. The new moment of inertia for the disk can be calculated using the parallel axis theorem:
I' = I + m * d^2
Where:
I' is the new moment of inertia,
I is the moment of inertia about the center of mass,
m is the mass of the disk,
d is the distance between the new axis and the center of mass.
In this case, the new axis is 1 cm away from the center, so d = 0.01 m.
Substituting the given values:
I' = (1/2) * 0.016 kg * (0.06 m)^2 + 0.016 kg * (0.01 m)^2
I' ≈ 3.648 × 10^(-5) kg·m²
Now, we can repeat the steps from part (a) to calculate the time it takes to bring the DVD to a stop:
τ = 0.5 N * 0.01 m
τ = 0.005 N·m
α = τ / I'
α = 0.005 N·m / 3.648 × 10^(-5) kg·m²
α ≈ 137.2 rad/s²
Δω = -1000 rpm * 2π rad/rev * 1 min/60 s
Δω = -1047.2 rad/s
Δt = Δω / α
Δt = (-1047.2 rad/s) / 137.2 rad/s²
Δt ≈ -7.63 s
The time it takes to bring the DVD to a stop when the friction force is applied 1 cm away from the center is approximately 7.63 seconds.
(c) The work done by the friction force to stop the wheel can be calculated using the equation:
Work = τ * θ
Where:
τ is the torque applied by the friction force,
θ is the angular displacement.
In both cases (a and b), the angular displacement is 180 degrees (π radians) since the DVD goes from rotating in one direction to not rotating at all.
For case (a), using the torque τ = 0.03 N·m:
Work = 0.03 N·m * π rad
Work ≈ 0.0942 J
For case (b), using the torque τ = 0.005 N·m:
Work = 0.005 N·m * π rad
Work ≈ 0.0157 J
Therefore, in case (a), the friction force does approximately 0.0942 Joules of work to stop the DVD, while in case (b), it does approximately 0.0157 Joules of work.