Lesson 21 – Centripetal Force: Tension Problems

1. Lesson 21– Centripetal Force: Tension Problems Fernando Morales August 28, 2013

2. Peer Instruction • A beginning physics student, confused by a seeming contradiction in Newton’s laws, asks her teacher the following question: “If, for every force there is an equal and opposite reaction force, then all forces in nature come in equal and opposite pairs, and are therefore balanced. Thus, since there can never be such a thing as an unbalanced force, how can any object ever accelerate?” Explain the fault in this common misconception.

3. Peer Instruction • A rodeo performer spins a lasso above his head. • Explain the purpose of twirling the rope before throwing it. • Describe how he could maximize the distance the rope can be thrown. • Describe the path the rope will take once he releases it.

4. Two Birds with One Stone • A brave warrior places a 1.0 kg rock in a sling and swings it in a horizontal circle around his head on a 1.0 m long vine. If the vine breaks at a tension of 200 N, what is the maximum speed, with which he can swing the rock? How would the answer change, if instead he used a vertical circle?

5. Peer Instruction • Rank in order, from largest to smallest, the centripetal accelerations (ar)a to (ar)e of particles a to e.

6. Tarzan • Tarzan swings on a vine that can only support 1,200N. The vine is 25m long and Tarzan has a mass of 90kg. What is the maximum speed that Tarzan can swing from one tree to the next and not have his vine break?

7. Sir Isaac Newton Contest Question • Jane, of mass 60 kg, stands on a 15 m high cliff holding a 12 m horizontal vine tied to a branch above the level plain below. She starts from rest and swings freely along the circular arc shown. When she reaches position 30 degrees below horizontal, she releases the vine. Calculate the tension in the vine at position II just before Jane let go. • Ans. T= 882 N

8. v² r v² r A ball moving in a circle a = According to Newton’s 2nd Law net force  acceleration Fnet = ma = m NB: this force often denoted Fc (for Centripetal Force) is a NET force – the actual force(s) involved might be tension in a string, gravity, electromagnetism or friction

9. A common misconception: many people think an object moving in a circle must have an outward force on it (after all, why do we feel as though we are crushed in our car seat when going around a corner if there is no outward force?) THIS IS INCORRECT – what we feel is the result of the conflict between our own inertia (we would continue moving in a straight line) and the force of the door pushing us around the corner An Outward Force? No The force on the object moving in a circle MUST be inward for it to go around the circle The force on the hand is outward but the force on the ball is inward

10. 4.     Tarzan plans to cross a gorge by swinging in an arc from a hanging vine.  If his arms are capable of exerting a force of 1500 N on the rope, what is the maximum speed he can tolerate at the lowest point of his swing?  His mass is 85 kg; the vine is 4.0 ml long.  v = 5.6m/s

11. 7.     Tarzan, at 96 kg, is swinging on a 9.3 m vine to reach Jane! At the bottom of the swing the tension in the vine is 1580 N. • a) What is Tarzan’s speed at the bottom of the swing? 7.87 m/s • b) What is Tarzan’s speed as he reaches Jane at an angle of 42º? 3.9 m/s • c) What is the tension in the vine just before he lets go and grabs onto Jane? 855 N

12. v ² v ² mv ² mg = r r v = rcos θ sin θ 2r T sin θ rg = mg mv ² cos θ v ² rcos θ sin θ rg tan θ = v² g T² tan θ = mg 4² r FT = sin 4 X5 - 3 EXAMPLE Force on a revolving ball A 0.250 kg ball on a 0.400 m string revolves horizontally at 3 revolutions per second. What is the centripetal force – the tension in the string, FT? Solution The cord sags at angle θ.Why? Because some force must balance mg down  a component of FT must be upward N2L: F = FT = ma = m X: FTcosθ = m Y: FTsinθ = mg FT = FT = T = 0.333 s (3 revs / s) = 0.0690 θ= 3.95º ~ 4° (sb rcos θ, not just r) = 35 N

13. X5 - 4 Tetherball When the ball is struck, it whirls around the pole. What is the direction of the acceleration, and what supplies that acceleration? EXAMPLE Solution The circle is horizontal  the acceleration is also horizontal (centripetal) The horizontal component of the tension in the string supplies the centripetal force

14. v ² r v ² r X5 - 5 Vertical Circle A 0.150 kg ball is swung in a vertical circle on a 1.10 m cord. (a) Find the minimum speed of the ball at the top so that it makes it around the circle. (b) Find the tension in the cord at the bottom of the circle if the ball is moving twice as fast as in (a) EXAMPLE Solution (a) The FBD at A shows mg and the tension in the same direction F = Fnet = mg + FT = m The speed will be minimum when FT goes to 0 mg = m v = (gr) = 3.28 m/s

15. v ² r v ² r X5 - 5 Vertical Circle A 0.150 kg ball is swung in a vertical circle on a 1.10 m cord. (a) Find the minimum speed of the ball at the top so that it makes it around the circle. (b) Find the tension in the cord at the bottom of the circle if the ball is moving twice as fast as in (a) EXAMPLE Solution (b) The FBD at B shows mg and the tension in opposite directions The speed is to be 2(3.28) = 6.56 m/s F = Fnet = FT – mg = m FT = m + mg = 7.34 N

17. Required Before Next Class • Chapter Review from Nelson Textbook # 31, 34, 35, 37, 38, 44, 48, 54