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Using the “Clicker”

Using the “Clicker”. If you have a clicker now, and did not do this last time, please enter your ID in your clicker. First, turn on your clicker by sliding the power switch, on the left, up. Next, store your student number in the clicker. You only have to do this once.

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Using the “Clicker”

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  1. Using the “Clicker” If you have a clicker now, and did not do this last time, please enter your ID in your clicker. First, turn on your clicker by sliding the power switch, on the left, up. Next, store your student number in the clicker. You only have to do this once. Press the * button to enter the setup menu. Press the up arrow button to get to ID Press the big green arrow key Press the T button, then the up arrow to get a U Enter the rest of your BU ID. Press the big green arrow key.

  2. Uniform Circular Motion • The path is a circle(radius r, circumference 2pr). • “Uniform” means constant speedv = 2pr / T, where the period T is the time to go around the circle once. • Angle in “radians” ≡ (arc length Ds) / (radius r) • Angular velocityw ≡ Dq/Dt = 2p/T[rad/sec], is also independent of r • Note thatv = (2p/T)r = w r[m/s], and therefore v is proportional to the radius of the circle. Dq = Ds1/r1 = Ds2/r2 is independent of the radius r of the circle, and is dimensionless

  3. Velocity on circular path Displacement for large time interval v = Dr/Dt but chord Dr is almost arc s = r Dq So again v= (rDq)/Dt = r(Dq/Dt) = wr = constant Displacement for small time interval Direction approaches tangent to circle, which is perpendicular to r For uniform circular motion, the velocity vector has magnitude v = wr, and direction is tangent to the circle at the position of the particle.

  4. Magnitude of the acceleration v2 v1 Dv For small time intervals, the vector Dv points toward the center, and has the magnitude Dv~ v Dq so a = Dv /Dt= v (Dq/Dt) = v w = v2/r -v1 Dq Dq v2 For uniform circular motion, the magnitude of the acceleration is w2r = v2/r, and the direction of the acceleration is toward the center of the circle.

  5. Coins on a turntable Two identical coins are placed on a flat turntable that is initially at rest. One coin is closer to the center than the other disk is. There is some friction between the coins and the turntable. We start spinning the turntable, steadily increasing the speed. Which coin starts sliding on the turntable first? • The coin closer to the center. • The coin farther from the center. • Neither, both coin start to slide at the same time.

  6. A general method for solving circular motion problems Follow the method for force problems! Draw a diagram of the situation. Draw one or more free-body diagrams showing all the forces acting on the object(s). Choose a coordinate system. It is often most convenient to align one of your coordinate axes with the direction of the acceleration. Break the forces up into their x and ycomponents. Apply Newton's Second Law in both directions. The key difference: usetoward the center

  7. Coins on a turntable (work together) Sketch a free-body diagram (side view) for one of the coins, assuming it is not sliding on the turntable. Apply Newton’s Second Law, once for each direction.

  8. Coins on a turntable (work together) FN Can you tell whether the velocity is into or out of the screen? FS Sketch a free-body diagram (side view) for one of the coins, assuming it is not sliding on the turntable. Axis of rotation mg

  9. Coins on a turntable (work together) FN y Can you tell whether the velocity is into or out of the screen? * FS x Apply Newton’s Second Law, once for each direction. y-direction: FN -mg = 0 so that FN = mg x-direction: FS = max = m(v2/r) [both FS and a are to left] Axis of rotation mg As you increase r, what happens to the force of friction needed to keep the coin on the circular path? * It is the same diagram and result either way!

  10. “Trick” question! v has a “hidden” dependence on r, so that the “obvious” dependence on r is not the whole story. The two coins have different speeds. Use angular velocity for the comparison, because the two coins rotate through the same angle in a particular time interval. This gives: As you increase r, what happens to the force of friction needed to keep the coin staying on the circular path? The larger r is, the larger the force of static friction has to be. The outer one hits the limit first.

  11. Conical pendulum A ball is whirled in a horizontal circle by means of a string. In addition to the force of gravity and the tension, which of the following forces should appear on the ball’s free-body diagram? A normal force, directed vertically up. A centripetal force, toward the center of the circle. A centripetal force, away from the center of the circle. Both 1 and 2. Both 1 and 3. None of the above.

  12. Conical pendulum (work together) Sketch a free-body diagram for the ball. Apply Newton’s Second Law, once for each direction.

  13. Conical pendulum (work together) Tsinq q q y Tcosq T Axis of rotation x Resolve mg Choose Sketch a free-body diagram for the ball. Apply Newton’s Second Law, once for each direction. y-direction: T cosq- mg = may= 0 x-direction: T sinq = max = m(v2/r) Solve: (mg/cosq)sinq = mv2/r (rg tanq )1/2 = v

  14. Gravitron (or The Rotor) In a particular carnival ride, riders are pressed against the vertical wall of a rotating ride, and then the floor is removed. Which force acting on each rider is directed toward the center of the circle? A normal force. A force of gravity. A force of static friction. A force of kinetic friction. None of the above.

  15. Gravitron (work together) FS He’s blurry because he is going so fast! Axis of rotation FN y mg x Sketch a free-body diagram for the rider. Apply Newton’s Second Law, once for each direction. y direction: FS- mg = may = 0 (he hopes) x direction: FN = max = m (v2/r)

  16. Test tonight • Go to COM 101. (Lecture section A1) • Test is 6-8 pm. • Test has more problems than I said, because some are shorter or easier. • “Best wishes!” (“Good luck” implies that you might not be fully prepared, and I don’t believe that for a minute.)

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