2018 USA Physics Olympiad Exam Part A 4 i.Suppose the sphere immediately begins to roll without slipping.What is the new speed of the sphere in this case? Solution If the sphere immediately begins to roll without slipping,we can calculate the frictional impulse independently of the normal impulse.We have nvsinθ-pF=mu. The frictional impulse is responsible for the sphere's rotation,so its angular momentum about its center of mass is L=pFR.But we also know that L BmR2w BmRu. Then PF Bmu. Substituting into the previous expression gives vsin0 mv sin0=(1+β)mu→ u=1+B ii.What is the minimum coefficient of friction such that the sphere rolls without slipping immediately after the collision? Solution As in part (a),the normal impulse is pN=mu cos0 and the maximal frictional impulse is pr =upN.From the previous part,we need Bmu sin 0 PF= 1+B and equating these expressions gives Btan0 μ= 1+B Copyright C2018 American Association of Physics Teachers2018 USA Physics Olympiad Exam Part A 4 i. Suppose the sphere immediately begins to roll without slipping. What is the new speed of the sphere in this case? Solution If the sphere immediately begins to roll without slipping, we can calculate the frictional impulse independently of the normal impulse. We have mv sin θ − pF = mu. The frictional impulse is responsible for the sphere’s rotation, so its angular momentum about its center of mass is L = pF R. But we also know that L = βmR2ω = βmRu. Then pF = βmu. Substituting into the previous expression gives mv sin θ = (1 + β)mu ⇒ u = v sin θ 1 + β . ii. What is the minimum coefficient of friction such that the sphere rolls without slipping immediately after the collision? Solution As in part (a), the normal impulse is pN = mv cos θ and the maximal frictional impulse is pF = µpN . From the previous part, we need pF = βmv sin θ 1 + β and equating these expressions gives µ = β tan θ 1 + β . Copyright c 2018 American Association of Physics Teachers