A string is wrapped around a uniform disk of mass M = 1.6 kg and radius R = 0.10 m. (Recall that the moment of inertia of a uniform disk is (1/2)MR2.) Attached to the disk are four low-mass rods of radius b = 0.17 m, each with a small mass m = 0.6 kg at the end. The device is initially at rest on a nearly frictionless surface. Then you pull the string with a constant force F = 23 N. At the instant when the center of the disk has moved a distance d = 0.047 m, a length w = 0.024 m of string has unwound off the disk.
You keep pulling with constant force 23 N for an additional 0.032 s. Now what is the angular speed of the apparatus?
To find the angular speed of the apparatus, we can use the law of conservation of angular momentum.
1. Calculate the initial angular momentum:
Initial angular momentum (L_initial) = Moment of inertia of the disk * initial angular velocity
The moment of inertia of the disk (I_disk) = (1/2) * M * R^2
The initial angular velocity (ω_initial) = 0 since the device is initially at rest.
So, L_initial = (1/2) * M * R^2 * 0
2. Calculate the final angular momentum:
Final angular momentum (L_final) = Moment of inertia of the disk * final angular velocity
The final angular velocity (ω_final) is what we need to find.
The length of string unwound (w) = 0.024 m
The distance moved by the center of the disk (d) = 0.047 m
The final angular displacement of the disk = w/d (since the string unwound comes from the circumference of the disk)
The angular displacement of the disk (θ) = ω_final * Δt
Δt = 0.032 s (time interval for pulling the string)
θ = (w/d) * Δt
L_final = (1/2) * M * R^2 * ω_final
3. Apply the law of conservation of angular momentum:
L_initial = L_final
(1/2) * M * R^2 * 0 = (1/2) * M * R^2 * ω_final
4. Solve for ω_final:
0 = ω_final
The final angular velocity is zero since there is no external torque acting on the system.
Therefore, the angular speed of the apparatus after an additional 0.032 s of pulling the string with a constant force of 23 N is zero.
To find the angular speed of the apparatus after pulling with a constant force for an additional 0.032 s, we need to calculate the change in angular displacement of the disk.
First, let's find the initial angular displacement of the disk when the center moved a distance d = 0.047 m. Since the length of the string unwound is w = 0.024 m, the arc length on the disk is also w. The arc length formula for a circular disk is given by:
s = θ * r
where s is the arc length, θ is the angular displacement in radians, and r is the radius.
Substituting the given values:
0.024 m = θ * 0.10 m
Now solve for θ:
θ = 0.024 m / 0.10 m
θ = 0.24 radians
Next, we need to find the final angular displacement after pulling with a constant force for an additional 0.032 s. The torque equation for rotational motion is given by:
τ = I * α
where τ is the torque, I is the moment of inertia, and α is the angular acceleration.
Since the force F is applied tangentially to the disk, it creates a torque. The torque generated by the force is given by:
τ = F * R
where R is the radius of the disk.
Substituting the given values:
τ = 23 N * 0.10 m
τ = 2.3 N*m
The torque is equal to the moment of inertia multiplied by the angular acceleration. The moment of inertia for the disk is (1/2)MR^2, so:
2.3 N*m = (1/2) * (1.6 kg) * (0.10 m)^2 * α
Now, solve for α:
α = (2.3 N*m) / [(1/2) * (1.6 kg) * (0.10 m)^2]
α = 57.5 rad/s^2
Now, we will use the kinematic equation relating angular displacement, initial angular velocity, angular acceleration, and time:
θ = ω_i * t + (1/2) * α * t^2
where θ is the final angular displacement, ω_i is the initial angular velocity, α is the angular acceleration, and t is the time.
Substituting the given values:
0.24 radians = ω_i * 0.032 s + (1/2) * (57.5 rad/s^2) * (0.032 s)^2
Now, solve for ω_i (initial angular velocity):
ω_i = (0.24 radians - (1/2) * (57.5 rad/s^2) * (0.032 s)^2) / (0.032 s)
ω_i = 0.24 radians / 0.032 s - (1/2) * (57.5 rad/s^2) * (0.032 s)
ω_i = 7.5 rad/s - 0.058 rad/s
ω_i = 7.442 rad/s
Therefore, the initial angular velocity of the apparatus is 7.442 rad/s.