Question

You have been appointed to an amusement ride safety committee for the Mall of America's Nickelodeon Universe, which is reviewing the safety of a ride that consists of seats mounted on each end of a rotating steel beam. For most of the ride, the beam rotates about its center in a horizontal circle at a constant speed. One committee member insists that a person moving in a circle at constant speed is not accelerating, so there is no need to be concerned about the ride’s safety. Another thinks that the person has a constant acceleration when moving at a constant speed. Yet a third argues that the person’s acceleration depends on the rate of change of their velocity, not their speed. Since each component of the person’s velocity changes with time, their acceleration must change with time. You decide to settle the issue by making a model of the ride and measuring the magnitude of the acceleration of different positions on the model when it spins at a constant speed.

1. Make a drawing of the path of an object in circular motion at constant speed. On that path, use a dot to represent the object’s position at time t 1 . Label this point as O, and draw a vector at O to represent the magnitude and direction of the object’s velocity at time t 1 . Draw another dot to represent the object’s position at a later time t 2 , shortly after t 1 , and label this point P. Draw a vector at P to show the magnitude and direction of the object’s velocity at time t 2 .

2. Redraw the velocity vectors with the tail of one vector (point P) at the tail of the other vector (point O). Keep the same size and direction as in the previous drawing. To find the acceleration of the object, you are interested in the change in velocity ( v). The change v is the increment that must be added to the velocity at time t 1 so that the resultant velocity has the new direction after the elapsed time t=t 2 – t 1 . Add the change in velocity v to your drawing of the velocity vectors; it should be a straight line connecting the heads of the vectors.

3. On your drawing from question 1, label the distance r from the center of the circle to points O and P. In the limit that the time interval is very small, the arc length distance traveled by the object can be approximated as a straight line. Use this approximation to label the distance traveled by the object along the circle from point O to P in terms of the object’s velocity and the elapsed time.

4. The triangle drawn in question 2 (with v and Δv) is similar to the triangle drawn in question 3 (with r and the straight line distance traveled by the object) because they have the same apex angle. Use the relationship of similar triangles to write an equation that connects the sides and the bases of the two triangles.

5. Solve your equation for Δv/Δt to get an expression for the acceleration in terms of the object’s uniform velocity and the distance r.

6. From your equation, is the acceleration of an object in circular motion ever zero? Does the magnitude of the acceleration change with time?

7. Does an object moving in a circle accelerate? If so, does the magnitude of the acceleration change with time? Explain your reasoning. Use the acceleration equation you derived in the Warm-up to support your claim.

Answer 1

Answer:- Given that, the appointed to an amusement side Safety Committed for the mall of America's Nickelodeon univers, which is reviewing the safely of a side that consists of seats mounted on each end of the rotating steel beam. (1) The path of an object in circular motion at constant speed. * label this point as o, and draw a vector at o and velocity time to, shortly after time to, and point po Draw a vector at p to show the magnitude and direction of the vector velocity object's time ta . tugees V & V are velocity of at time t, & to object lil = v. & 1 l = 42 (2) The velocity vectors with the tail of one vector at the tail of point o, velocity & time t, and time t = ta-ti.

AV = V V charge in velocity DV = OP = 72 72 (3) from this it o is small, are length is straight line => tano = . =) or artano since o į small tan o i distance travelled from o top Ar = 80 (4) The triangle drawn in v and ar is similar triangle the equation that have same apex angle the sides and bases of the two triangls.

ale À op & top are is transfer TV FILM => or = ☆ Byl At and ato, then lim ora sy l lim AV - za At- 4o at a az Ý x V = V2 © Since o&v are constant ia is also constant. It does not charge with time. 'a' is not equal to o but a= ut Cour 0 yes object moving is a circle accelerate this is becaus althrough object moves with same speed, but every in direction. of velocity changes. Hence object moving į a circle acceleration. ance

Magnitude of acceleration is given by 2 i accelaration is not zejoy and also Since are constand hencé 'á is also constant. v fx