If A Uniform-density Disk Of Mass 5 Kg And Radius 0.2 M Is Rotating About Its Central Axis With An Angular Velocity Of 3 Rad/s, And A Point Mass Of 2 Kg Is Attached To The Disk's Rim, What Is The New Moment Of Inertia Of The System, And How Will The Angular Velocity Change If The Point Mass Is Released From Rest At A Distance Of 0.15 M From The Disk's Center, Assuming No External Torques Are Applied?

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First, we calculate the moment of inertia of the uniform-density disk, which is given by the formula I=12MR2{ I = \frac{1}{2}MR^2 }. For the disk with mass 5 kg and radius 0.2 m, the moment of inertia is:

Idisk=12×5×(0.2)2=0.1 kg\cdotpm2{ I_{\text{disk}} = \frac{1}{2} \times 5 \times (0.2)^2 = 0.1 \text{ kg·m}^2 }

When a point mass of 2 kg is attached to the rim of the disk, its moment of inertia is calculated as I=MR2{ I = MR^2 }. For the point mass at a radius of 0.2 m, the moment of inertia is:

Ipoint=2×(0.2)2=0.08 kg\cdotpm2{ I_{\text{point}} = 2 \times (0.2)^2 = 0.08 \text{ kg·m}^2 }

The total moment of inertia of the system when the point mass is attached to the rim is:

Itotal=Idisk+Ipoint=0.1+0.08=0.18 kg\cdotpm2{ I_{\text{total}} = I_{\text{disk}} + I_{\text{point}} = 0.1 + 0.08 = 0.18 \text{ kg·m}^2 }

Next, we consider the scenario where the point mass is released from rest at a distance of 0.15 m from the disk's center. The new moment of inertia of the point mass at this distance is:

Ipoint′=2×(0.15)2=0.045 kg\cdotpm2{ I'_{\text{point}} = 2 \times (0.15)^2 = 0.045 \text{ kg·m}^2 }

The total moment of inertia of the system after the point mass is released is:

Itotal′=Idisk+Ipoint′=0.1+0.045=0.145 kg\cdotpm2{ I'_{\text{total}} = I_{\text{disk}} + I'_{\text{point}} = 0.1 + 0.045 = 0.145 \text{ kg·m}^2 }

Using the conservation of angular momentum, we know that the initial angular momentum L1=I1ω1{ L_1 = I_1 \omega_1 } must equal the final angular momentum L2=I2ω2{ L_2 = I_2 \omega_2 }. The initial angular momentum is:

L1=0.18×3=0.54 kg\cdotpm2/s{ L_1 = 0.18 \times 3 = 0.54 \text{ kg·m}^2/\text{s} }

Setting this equal to the final angular momentum:

0.54=0.145×ω2{ 0.54 = 0.145 \times \omega_2 }

Solving for ω2{ \omega_2 }:

ω2=0.540.145=10829≈3.724 rad/s{ \omega_2 = \frac{0.54}{0.145} = \frac{108}{29} \approx 3.724 \text{ rad/s} }

Thus, the new moment of inertia of the system is \boxed{0.18} kg·m², and the angular velocity changes to \boxed{\frac{108}{29}} rad/s.