A solid disk of mass m1 = 9.4 kg and radius R = 0.25 m is rotating with a constant angular velocity of ω = 36 rad/s. A thin rectangular rod with mass m2 = 3.1 kg and length L = 2R = 0.5 m begins at rest above the disk and is dropped on the disk where it begins to spin with the disk.
a. What is the initial angular momentum of the rod and disk system?
b. What is the initial rotational energy of the rod and disk system?
c. What is the final angular velocity of the disk?
d. What is the final angular momentum of the rod and disk system?
e. What is the final rotational energy of the rod and disk system?
f. The rod took t = 5.7 s to accelerate to its final angular speed with the disk. What average torque was exerted on the rod by the disk?
a. The initial angular momentum of the rod and disk system is L = m1R2ω = 6.9 kg m2/s.
b. The initial rotational energy of the rod and disk system is Erot = ½Iω2 = ½(m1R2 + m2L2)ω2 = 4.2 J.
c. The final angular velocity of the disk is ω = 36 rad/s.
d. The final angular momentum of the rod and disk system is L = m1R2ω + m2L2ω = 13.8 kg m2/s.
e. The final rotational energy of the rod and disk system is Erot = ½Iω2 = ½(m1R2 + m2L2)ω2 = 16.2 J.
f. The average torque exerted on the rod by the disk is τ = (Lf - Li)/t = (13.8 - 6.9)/5.7 = 1.4 Nm.
To solve this problem, we'll need to use the principles of angular momentum and conservation of energy.
a) The initial angular momentum of the rod and disk system can be calculated by adding the angular momentum of the disk and the angular momentum of the rod.
The angular momentum of the disk is given by L_disk = I_disk * ω, where I_disk is the moment of inertia of the disk. For a solid disk rotating about its axis, the moment of inertia is I_disk = (1/2) * m1 * R^2.
The angular momentum of the rod is given by L_rod = I_rod * ω, where I_rod is the moment of inertia of the rod. For a thin rectangular rod rotating about its center perpendicular to its length, the moment of inertia is I_rod = (1/12) * m2 * L^2.
So, the initial angular momentum of the system is L_initial = L_disk + L_rod.
b) The initial rotational energy of the rod and disk system can be calculated using the formula E_initial = (1/2) * I_disk * ω^2 + (1/2) * I_rod * ω^2.
c) When the rod is dropped onto the disk and starts spinning with it, the law of conservation of angular momentum tells us that the total angular momentum of the system remains constant.
The final angular velocity of the disk can be calculated using the equation L_initial = L_final, where L_final is the angular momentum of the system after the rod is dropped onto the disk. Rearranging the equation gives ω_final = L_initial / I_disk.
d) The final angular momentum of the rod and disk system can be calculated in the same way as the initial angular momentum: L_final = L_disk + L_rod.
e) The final rotational energy of the rod and disk system can be calculated using the formula E_final = (1/2) * I_disk * ω_final^2 + (1/2) * I_rod * ω_final^2.
f) The average torque exerted on the rod by the disk can be calculated using the equation τ = ΔL / Δt, where τ is the torque, ΔL is the change in angular momentum of the rod, and Δt is the time interval.
To find ΔL, we subtract the initial angular momentum of the rod from the final angular momentum: ΔL = L_final - L_rod. The time interval Δt is given as 5.7 seconds.
Finally, we can calculate the average torque τ = ΔL / Δt.
To solve this problem, we can use the conservation of angular momentum and energy.
a. The initial angular momentum (L) of the rod and disk system is given by the sum of the individual angular momenta:
L1 = I1 * ω1 (for the disk)
L2 = I2 * ω2 (for the rod)
Since the disk is rotating at a constant angular velocity (ω1 = 36 rad/s), its initial angular momentum (L1) is:
L1 = I1 * ω1 = (1/2) * m1 * R^2 * ω1 = (1/2) * 9.4 kg * (0.25 m)^2 * 36 rad/s = ¼ * 9.4 kg * 0.25 m^2 * 36 rad/s = 31.5 kg·m^2/s
Since the rod is at rest initially, its initial angular momentum (L2) is zero:
L2 = 0
Therefore, the initial angular momentum of the rod and disk system is:
L = L1 + L2 = 31.5 kg·m^2/s + 0 = 31.5 kg·m^2/s
b. The initial rotational energy (E) of the rod and disk system is given by the sum of the individual rotational energies:
E1 = (1/2) * I1 * ω1^2 (for the disk)
E2 = (1/2) * I2 * ω2^2 (for the rod)
The initial rotational energy of the disk (E1) is:
E1 = (1/2) * I1 * ω1^2 = (1/2) * 9.4 kg * (0.25 m)^2 * (36 rad/s)^2 = 39.96 J
The initial rotational energy of the rod (E2) is zero, since it is initially at rest:
E2 = 0
Therefore, the initial rotational energy of the rod and disk system is:
E = E1 + E2 = 39.96 J + 0 = 39.96 J
c. The final angular velocity (ωf) of the disk can be determined using the principle of conservation of angular momentum:
L = I * ωf
Since the rod is dropped on the disk and starts spinning with the disk, the final angular momentum (Lf) of the rod and disk system remains the same as the initial angular momentum (L):
Lf = L = 31.5 kg·m^2/s
The moment of inertia (I) of the disk is given by:
I = (1/2) * m1 * R^2 = (1/2) * 9.4 kg * (0.25 m)^2 = 0.5875 kg·m^2
Therefore, the final angular velocity of the disk (ωf) is:
Lf = I * ωf
ωf = Lf / I = 31.5 kg·m^2/s / 0.5875 kg·m^2 = 53.67 rad/s (approximately)
The final angular velocity of the disk is approximately 53.67 rad/s.
d. The final angular momentum (L) of the rod and disk system is the same as the initial angular momentum (L):
L = 31.5 kg·m^2/s
e. The final rotational energy (E) of the rod and disk system can be calculated using the final angular velocity (ωf) and the moment of inertia (I) of the disk:
E = (1/2) * I * ωf^2
E = (1/2) * 0.5875 kg·m^2 * (53.67 rad/s)^2
E ≈ 848.6 J
The final rotational energy of the rod and disk system is approximately 848.6 J.
f. The average torque (τ) exerted on the rod by the disk can be calculated using the equation:
τ = ΔL / Δt
Where ΔL is the change in angular momentum and Δt is the time interval.
Since the rod took t = 5.7 s to accelerate to its final angular speed with the disk, the change in angular momentum (ΔL) is:
ΔL = Lf - L = L - L = 0
Therefore, the average torque exerted on the rod by the disk is:
τ = ΔL / Δt = 0 / 5.7 s = 0 N·m