Question

4)A large disk of mass M_d and radius R is spinning in a horizontal plane about a vertical axis through its center with a given angular velocity of omega. A person of mass M_p (initially at the center of the disk and not spinning, let's say) then walks out to the edge of the disk (yes, the disk is that large!). Find the final angular velocity of the disk (with the person standing on its edge).Treat the person as a point mass. Show step by step and Show a Diagram thank you(Picture)

Answer 1

Answer:

let ω¹ be the final angular velocity

using conservation of momentum,

we have

M_dR²/2 x ω = (M_pR² + M_dR²/2 ) ω¹

so,

ω¹ =

Therefore, the answer is

Answer 2

Key Concept: Conservation of Angular Momentum

Since no external torque acts on the system (disk + person), the total angular momentum ($L$) remains constant:

$$L_{\text{initial}}=L_{\text{final}}$$

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Step 1: Calculate Initial Angular Momentum ($L_i$)

1. Disk's contribution:
   Moment of inertia of a disk:

   $$I_d=\frac{1}{2}{M}_d{R}^2$$

   Angular momentum of the disk:

   $$L_d=I_d\omega =\frac{1}{2}{M}_d{R}^2\omega$$

2. Person's contribution (at center):
   Since the person is at the center ($r=0$), their moment of inertia is:

   $$I_p=M_p\cdot 0^2=0$$

   So, their initial angular momentum is zero.

3. Total initial angular momentum:

   $$L_i=L_d+L_p=\frac{1}{2}{M}_d{R}^2\omega +0=\frac{1}{2}{M}_d{R}^2\omega$$

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Step 2: Calculate Final Angular Momentum ($L_f$)

1. Disk's contribution (unchanged):

   $$I_d=\frac{1}{2}{M}_d{R}^2$$$$L_d=I_d{\omega }_f=\frac{1}{2}{M}_d{R}^2{\omega }_f$$

2. Person's contribution (now at edge):
   Moment of inertia of the person (treated as a point mass at radius $R$):

   $$I_p=M_p{R}^2$$

   Angular momentum of the person:

   $$L_p=I_p{\omega }_f=M_p{R}^2{\omega }_f$$

3. Total final angular momentum:

   $$L_f=L_d+L_p=\frac{1}{2}{M}_d{R}^2{\omega }_f+M_p{R}^2{\omega }_f$$

   Factor out $R^2{\omega }_f$:

   $$L_f=(\frac{1}{2}{M}_d+M_p)R^2{\omega }_f$$

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Step 3: Apply Conservation of Angular Momentum ($L_i=L_f$)

$$\frac{1}{2}{M}_d{R}^2\omega =(\frac{1}{2}{M}_d+M_p)R^2{\omega }_f$$

Divide both sides by $R^2$ (since $R\mathrm{\ne }0$):

$$\frac{1}{2}{M}_d\omega =(\frac{1}{2}{M}_d+M_p){\omega }_f$$

Solve for ${\omega }_f$:

$${\omega }_f=\frac{\frac{1}{2}{M}_d\omega }{\frac{1}{2}{M}_d+M_p}$$

Multiply numerator and denominator by 2 to simplify:

$${\omega }_f=\frac{M_d\omega }{M_d+2M_p}$$

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Final Answer:

$$\boxed{{\displaystyle {\omega }_f=\left(\frac{M_d}{M_d+2M_p}\right)\omega }}$$