Mini-Exam 4 - PHY2048C - November 7, 2024 PDF

PHY2048C 1 of 8 Code: 163500 Mini-Exam 4 November 7, 2024 Name (write out fully in BLOCK letters): ___________________________________________...

PHY2048C 1 of 8 Code: 163500 Mini-Exam 4 November 7, 2024 Name (write out fully in BLOCK letters): _____________________________________________________________________________________ By signing, I acknowledge that I understand and will follow the FSU Honor Code (cheating means a zero on the test). The test will not be graded without a signature. Signature: _____________________________________________________________________________________ The answers need to be entered on the Scantron sheet provided, together with the “Code” for the test in the “SPECIAL CODING SECTION” on the Scantron and your EMPLID in the “IDENTIFICATION NO.” section. Do not forget to also enter your name (last name rst), your instructor’s name, the course number, and the date. The “Form” eld is not used. Your grade will be based entirely on what is entered in the Scantron sheets. The exam books will not be graded and are only used to work out the problems and as a cross-check in case of problems.   fi fi  PHY2048C 2 of 8 Code: 163500 This page intentionally left blank    PHY2048C 3 of 8 Code: 163500 1. Two circular gears of radius R and 2R are connected by interlocking teeth as shown in the diagram. If the larger gear spins at an angular velocity ω, what is the angular velocity of the smaller gear? (a) 4ω 1 (b) ω 2 (c) 2ω 1 (d) ω 4 (e) ω Chapter 10 | Rotational Motion and Angular Momentum Making Connections In statics, the net torque is zero, and there is no angular acceleration. In rotati acceleration, exactly as in Newton’s second law of motion for rotation. 2. An ice skater spins with his arms fully extended at an angular speed ω. Each arm has length L, and the skater’s moment of inertia can be modeled as that of a rod pivoted about its center of mass. If the skater folds his arms such that they each now extend out a distance L /2, what will be his new angular speed? (a) ω /4 (b) 4ω (c) ω (d) ω /2 (e) 2ω Figure 10.12 Some rotational inertias.                    PHY2048C 4 of 8 Code: 163500 3. A uniform rod of length L and mass M is attached to a wall via a frictionless pivot as shown in the diagram. The rod is released from rest from a horizontal position. What is its initial angular acceleration? 3 g (a) 2L g (b) L g (c) 6 L 1 g (d) 2L g (e) 3 L 4. A uniform solid sphere is released from rest from an incline of height h and angle θ. Assuming it rolls without slipping, what is the sphere’s velocity at the bottom of the incline? 4 (a) gh 3 24 (b) gh 13 10 (c) gh 7 (d) gh (e) 2gh                 PHY2048C 5 of 8 Code: 163500 5. A wad of clay of mass M /10 and velocity v is thrown horizontally at a solid uniform cylinder of mass M and radius R initially at rest. The cylinder is free to rotate about a xed axle through its center. The wad of clay sticks to the outer rim of the cylinder at the top, as indicated in the gure, causing the clay and cylinder to start to rotate. What is the angular speed of the cylinder just after the collision? 2 v (a) 11 R 1 v (b) 11 R 1 v (c) 6R 20 g (d) 11 v 20 v (e) 11 R 6. A constant horizontal force F is applied to a lawn roller that is a uniform solid cylinder of mass M by pulling perpendicular to an axis through its center as shown in the diagram. What is the acceleration of the center of mass? 2 F (a) 3M 4 F (b) 3M F (c) M 1 F (d) 3M 1 F (e) 2M              fi fi       PHY2048C 6 of 8 Code: 163500 7. A sign of mass M hangs from the end of a rod of length L and mass m supported by a cable as shown in the gure. The mass distributions in the sign and the rod are uniform. What is the tension in the cable? m g (a) ( +M) 2 cos θ m g (b) ( +M) 2 sin θ Mg (c) cos θ g (d) (m + M ) cos θ Mg (e) sin θ 8. A uniform board of length L and mass M is supported by a fulcrum placed a distance L /3 from the left end of the board. If a mass M /3 is placed on the left end of the board, at what distance from the fulcrum should a counterweight of the same mass M /3 be placed to balance the board? 5 (a) L to the right of the fulcrum 9 1 (b) L to the right of the fulcrum 6 1 (c) L to the left of the fulcrum 6 1 (d) L to the right of the fulcrum 3 1 (e) L to the left of the fulcrum 3               fi       PHY2048C 7 of 8 Code: 163500 Equations F i⃗ ≡ F net ⃗ = m a,⃗ i runs over all forces acting on the body ∑ i g = 9.8 m/s2 mv 2 Common forces: Fcentripetal = , w = mg, Ffric, static = μs N, Ffric, kinetic = μk N, R N ≡ | N ⃗ | = magnitude of the normal force 1 1 2 1 2 Energy: K Etrans = mv 2, K Erot = Iω , PEgrav = mgh, PEspring = kx 2 2 2 Work: W = F ⃗ ⋅ d ⃗ = Fd cos θ Work-energy theorem: W = ΔK E Conservation of energy: Wnon−conservative + K Ei + PEi = K Ef + PEf Linear momentum: p⃗ = m v⃗ (conserved when no net force acts on a system) W Average power: P̄ = Δt v a Rotational motion: ω = , α= R r Torque: τ ⃗ = r ⃗ × F ,⃗ τ = rF sin θ = Iα ⃗ = I α ,⃗ i runs over all torques acting on the body ∑ τi⃗ ≡ τnet i Angular momentum: L ⃗ = r ⃗ × p⃗ = I ω ⃗ (conserved when no net torque acts on a system)                          pter 10 | Rotational Motion and Angular Momentum 35 PHY2048C 8 of 8 Code: 163500 Point mass: I = MR 2 aking Connections statics, the net torque is zero, and there is no angular acceleration. In rotational motion, net torque is the cause of angular cceleration, exactly as in Newton’s second law of motion for rotation. gure 10.12 Some rotational inertias. Example 10.7 Calculating the Effect of Mass Distribution on a Merry-Go-Round onsider the father pushing a playground merry-go-round in Figure 10.13. He exerts a force of 250 N at the edge of the 0.0-kg merry-go-round, which has a 1.50 m radius. Calculate the angular acceleration produced (a) when no one is on the erry-go-round and (b) when an 18.0-kg child sits 1.25 m away from the center. Consider the merry-go-round itself to be a niform disk with negligible retarding friction.