© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Forces and Motions
When the hoop is placed on a rough horizontal surface with an initial angular velocity $\omega_0$ and its center of mass initially at rest, friction acts at the point of contact. This friction accelerates the hoop’s center of mass forward and also exerts a torque that reduces the angular velocity.
Step 2: Write Down the Equations of Motion
Let $v$ be the linear velocity of the center of the hoop and $\omega$ be the angular velocity at any instant after being placed on the surface. Let $f$ be the frictional force.
1. Translational motion: mdvdt=f Thus, dvdt=fm
2. Rotational motion (about the center of the hoop): Idωdt=-fr For a hoop, the moment of inertia about its center is $I = m r^2$. Hence, mr2dωdt=-fr ⟹ dωdt=-fmr
Step 3: Relate the Changes in Linear and Angular Velocities
We want the condition when the hoop just ceases to slip, meaning the linear speed $v$ becomes related to the angular speed $\omega$ by the pure rolling condition: v=r ω.
To find $v$ and $\omega$ at that instant, notice that $v$ increases from $0$ to $v_{\text{final}}$ and $\omega$ decreases from $\omega_0$ to $\omega_{\text{final}}$. We can use the ratio of the differentials: dvdt=fm, dωdt=- fmr. Dividing the first by the second gives dvdω=fm-fmr=- r. Integrating from the initial to final values: ∫0vfinaldv=-r∫ω0ωfinaldω. This simplifies to vfinal-0=-r\bigl(ωfinal-ω0\bigr).
Step 4: Apply the No-Slip (Rolling) Condition
At the instant slipping ceases, $v_{\text{final}} = r\,\omega_{\text{final}}$. From our integral result: vfinal=r\bigl(ω0-ωfinal\bigr). Therefore, setting $v_{\text{final}} = r\,\omega_{\text{final}}$ gives r ωfinal=r\bigl(ω0-ωfinal\bigr). Simplifying, ωfinal=ω0-ωfinal, 2 ωfinal=ω0, ωfinal=ω02. Substituting back into the rolling condition $v_{\text{final}} = r\,\omega_{\text{final}}$, we get: vfinal=r×ω02=r ω02.
Step 5: Conclude the Result
Hence, when the hoop ceases to slip, the velocity of its center of mass is r ω02.
