Question

2. As Wile E continues giving chase, he finds himself skating on ice. As the road runner whooshes past him, he spins with an angular speed of 12.0 rad/s. By opening up his arms, Wile E. increases his moment of inertia to double its initial value. What is Wile E.’s final angular speed?

Please Show All Work

Answer 1

Answer 2

Solution:

This problem involves the conservation of angular momentum, which states:

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

Angular momentum ($L$) is given by:

$$L=I\omega$$

where:

• $I$ = moment of inertia,

• $\omega$ = angular speed.

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Given:

• Initial angular speed (${\omega }_i$) = 12.0 rad/s,

• Final moment of inertia ($I_f$) = $2\times I_i$ (double the initial).

Step 1: Apply Conservation of Angular Momentum

$$I_i{\omega }_i=I_f{\omega }_f$$

Substitute $I_f=2I_i$:

$$I_i\cdot 12.0=2I_i\cdot {\omega }_f$$

Step 2: Solve for Final Angular Speed (${\omega }_f$)

$$12.0=2{\omega }_f⟹{\omega }_f=\frac{12.0}{2}=6.0\text{rad/s}$$

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Answer:

Wile E.’s final angular speed is 6.0 rad/s.