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Topic 2: Mechanics 2.4 Uniform circular motion

2.4.1 Draw a vector diagram to illustrate that the acceleration of a particle moving with constant speed in a circle is directed towards the center of the circle. 2.4.2 Apply the expressions for centripetal acceleration.

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Topic 2: Mechanics 2.4 Uniform circular motion

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  1. 2.4.1 Draw a vector diagram to illustrate that the acceleration of a particle moving with constant speed in a circle is directed towards the center of the circle. 2.4.2 Apply the expressions for centripetal acceleration. 2.4.3 Identify the force producing circular motion in various situations. Examples include friction of tires on turn, gravity on planet/moon, cord tension. 2.4.4 Solve problems involving circular motion. Topic 2: Mechanics2.4 Uniform circular motion

  2. r v On the next pass, however, Helen failed to clear the mountains. Topic 2: Mechanics2.4 Uniform circular motion What force must be applied to Helen to keep her moving in a circle? How does it depend on the Helen’s radius r? How does it depend on Helen’s velocity v? How does it depend on Helen’s mass m? m

  3. y v red x r blue Draw a vector diagram to illustrate that the acceleration of a particle moving with constant speed in a circle is directed towards the center of the circle. A particle is said to be in uniform circular motion if it travels in a circle (or arc) with constant speed v. Observe that the velocity vector is always tangent to the circle. Note that the magnitude of the velocity vector is NOT changing. Note that the direction of the velocity vector IS changing. Thus, there is an acceleration, even though the speed is not changing! Topic 2: Mechanics2.4 Uniform circular motion v r

  4. y v red x r blue v2 v2 v1 -v1 v1 ∆v Draw a vector diagram to illustrate that the acceleration of a particle moving with constant speed in a circle is directed towards the center of the circle. In order to find the direction of the acceleration (a = ∆v/∆t) we observe two nearby snapshots of the particle: The direction of the acceleration is gotten from ∆v = v2– v1 = v2+ (-v1): The direction of the acceleration is toward the center of the circle- you must be able to sketch this. Topic 2: Mechanics2.4 Uniform circular motion v2 v1 -v1 ∆v

  5. Fc =mac centripetal force Fc Apply the expressions for centripetal acceleration. Centripetal means center-seeking. How does centripetal acceleration acdepend on r and v? We define the centripetal forceFc: Picture yourself as the passenger in a car that is rounding a left turn: The sharper the turn, the harder you and your door push against each other. (Small r = big Fc.)  The faster the turn, the harder you and your door push against each other. (Big v = big Fc.) Topic 2: Mechanics2.4 Uniform circular motion

  6. manipulated No change r r r r ac ac ac ac no change manipulated v v v v responding responding Fc Fc Fc Fc Apply the expressions for centripetal acceleration. Topic 2: Mechanics2.4 Uniform circular motion PRACTICE: For each experiment A and B, label the control, independent, and dependent variables. CONTROL: r CONTROL: v INDEPENDENT: v INDEPENDENT: r A B DEPENDENT: Fc and ac DEPENDENT: Fc and ac

  7. first guess formula ac =v2/r centripetal acceleration Apply the expressions for centripetal acceleration. We know the following things about ac: If v increases, ac increases. If r increases, ac decreases. From dimensional analysis we have What can we do to v or r to “fix” the units? This is the correct one! Topic 2: Mechanics2.4 Uniform circular motion v r ac= ? ? v r  1 s m s2 m/s m ac= = = ? ?  m s2 m2/s2 m m s2 v2 r = = ac=

  8. Fc =mac centripetal force ac =v2/r centripetal acceleration Apply the expressions for centripetal acceleration. Topic 2: Mechanics2.4 Uniform circular motion • EXAMPLE: A 730-kg Smart Car negotiates a 30. m radius turn at 25. ms-1. What is its centripetal acceleration and force? What force is causing this acceleration? • SOLUTION: • ac =v2/r= 252/30 = 21 ms-2. • Fc =mac= (730)(21) = 15000 n. • The centripetal force is caused by the friction force between the tires and the pavement.

  9. ac =42r/T2 ac =v2/r centripetal acceleration Apply the expressions for centripetal acceleration. The periodT is the time for one complete revolution. One revolution is one circumference C = 2r. Therefore v = distance / time = 2r/T. Thus v2 = 42r2/T2 so that ac =v2/r = 42r2/T2r =42r/T2. Topic 2: Mechanics2.4 Uniform circular motion

  10. ac =42r/T2 ac =v2/r centripetal acceleration Apply the expressions for centripetal acceleration. Topic 2: Mechanics2.4 Uniform circular motion • EXAMPLE: Albert the 2.50-kg physics cat is being swung around your head by a string harness having a radius of 3.00 meters. He takes 5.00 seconds to complete one fun revolution. What are ac and Fc? What is the tension in the string? • SOLUTION: • ac =42r/T2 • = 42(3)/(5)2 = 4.74 ms-2. • Fc =mac = (2.5)(4.74) = 11.9 n. The tension is causing the centripetal force, so the tension is Fc = 11.9 n. Albert the Physics Cat

  11. A B C Solve problems involving circular motion. EXAMPLE: Dobson is watching a 16-pound bowling ball being swung around at 50 m/s by Arnold. If the string is cut at the instant the ball is next to the ice cream, what will the ball do? (a)It will follow path A and strike Dobson's ice cream. (b)It will fly outward along curve path B. (c)It will fly tangent to the original circular path along C. Topic 2: Mechanics2.4 Uniform circular motion

  12. T 1.50 m W Solve problems involving circular motion. EXAMPLE: A 3.0-kg mass is tied to a string having a length of 1.5 m, and placed in uniform circular motion as shown. The string traces out a cone with a base angle of 60°, with the mass traveling the base of the cone. (a) Sketch in the forces acting on the mass. SOLUTION: The ONLY two forces acting on the mass are its weight W and the tension in the string T. Don’t make the mistake of drawing Fc into the diagram. Fc is the resultant of T and W, as the next questions will illustrate. Topic 2: Mechanics2.4 Uniform circular motion

  13. T T Ty 1.50 m Tx W W Solve problems involving circular motion. EXAMPLE: A 3.0-kg mass is tied to a string having a length of 1.5 m, and placed in uniform circular motion as shown. The string traces out a cone with a base angle of 60°, with the mass traveling the base of the cone. (b) Draw a FBD in the space provided. Then break down the tension force in terms of the unknown tension T. SOLUTION: Tx = Tcos = T cos60° = 0.5T. Ty = Tsin = T sin60° = 0.87T. Topic 2: Mechanics2.4 Uniform circular motion 

  14. T Ty Tx W Solve problems involving circular motion. EXAMPLE: A 3.0-kg mass is tied to a string having a length of 1.5 m, and placed in uniform circular motion as shown. The string traces out a cone with a base angle of 60.°, with the mass traveling the base of the cone. (c) Find the value of the components of the tension Txand Ty. SOLUTION: Note that Ty = mg = 3.0(9.8) = 29.4 n. But Ty = 0.87T. Thus 29.4n = 0.87T so that T = 33.8 n. Tx = 0.5T =0.5(46) = 23 n. Topic 2: Mechanics2.4 Uniform circular motion

  15. 1.50 m r  Solve problems involving circular motion. EXAMPLE: A 3.0-kg mass is tied to a string having a length of 1.5 m, and placed in uniform circular motion as shown. The string traces out a cone with a base angle of 60°, with the mass traveling the base of the cone. (d) Find the speed of the mass as it travels in its circular orbit. SOLUTION: The center of the UCM is here: Since r / 1.5 = cos60°,r= 0.75 m. Fc = Tx = 23 n. Thus Fc = mv2/r 23 = 3v2/0.75 v = 2.4 ms-1. Topic 2: Mechanics2.4 Uniform circular motion

  16. Solve problems involving circular motion. EXAMPLE: Suppose a 0.500-kg baseball is placed in a circular orbit around the earth at slightly higher that the tallest point, Mount Everest (8850 m). Given that the earth has a radius of RE = 6400000 m, find the speed of the ball. SOLUTION: The ball is traveling in a circle of radius r = 6408850 m. Fc is caused by the weight of the ball so that Fc = mg = (0.5)(10) = 5 n. Since Fc = mv2/rwe have 5 = (0.5)v2/6408850 v = 8000 ms-1! Topic 2: Mechanics2.4 Uniform circular motion

  17. Solve problems involving circular motion. EXAMPLE: Suppose a 0.500-kg baseball is placed in a circular orbit around the earth at slightly higher that the tallest point, Mount Everest (8850 m). How long will it take the ball to return to Everest? SOLUTION: We want to find the period T. We know that v = 8000 ms-1. We also know that r= 6408850 m. Since v = 2r/T we have T = 2r/v T = 2(6408850)/8000 T = (5030 s)(1 h / 3600 s) = 1.40 h. Topic 2: Mechanics2.4 Uniform circular motion

  18. Solve problems involving circular motion. EXAMPLE: Explain how an object can remain in orbit yet always be falling. SOLUTION: Throw the ball at progressively larger speeds. In all instances the force of gravity will draw the ball toward the center of the earth. When the ball is finally thrown at a great enough speed, the curvature of the ball’s path will match the curvature of the earth’s surface. The ball is effectively falling around the earth! Topic 2: Mechanics2.4 Uniform circular motion

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