Irodov Problem 1.167 – Ball Thrown with Backspin

May 17, 2026

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Problem Statement

A ball of radius R is thrown horizontally with speed v₀ and backspin ω₀ (ω₀R = v₀). Find the subsequent motion on a rough floor (friction μ).

Given Information

v₀ forward, ω₀ backspin, ω₀R = v₀
Floor friction μ

Physical Concepts & Formulas

Friction decelerates translation and changes rotation.

Step-by-Step Solution

Step 1: Initial contact point moves backward at v₀ − ω₀R = 0… but backspin means bottom of ball moves forward (same as ball) but oppositely — contact point moves at v₀ + ω₀R = 2v₀ forward.

Step 2: Friction acts backward: a = −μg, α = +μmgR/I (reducing backspin).

Step 3: Rolling begins when v = ωR, giving v_f = v₀/3 (similar to previous problem).

Worked Calculation

Final rolling velocity = v₀/3 (for solid sphere with appropriate I)

Answer

$$\boxed{v_{final}=\frac{v_0}{3}\text{ (for solid sphere)}}$$

Physical Interpretation

Backspin initially means friction acts backward, slowing the ball. Rolling begins when translational and rotational speeds become compatible. This is why pool balls with backspin can stop and reverse.