physics ch6 outline

physics ch6 outline - Chapter 6 Circular Motion and Other...

Info icon This preview shows pages 1–10. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 6 Circular Motion and Other Applications of Newton’s Laws
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Uniform Circular Motion Uniform circular motion occurs when an object moves in a circular path with a constant speed The velocity vector is always tangent to the path of the object
Image of page 2
Changing Velocity in Uniform Circular Motion The change in the velocity vector is due to the change in direction The vector diagram shows f i = + ∆ v v v r r r
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Centripetal Acceleration The acceleration is always perpendicular to the path of the motion and always points toward the center of the circle of motion This acceleration is called the centripetal acceleration. The magnitude of the centripetal acceleration vector is given by 2 C v a r =
Image of page 4
Period The period , T , is the time required for one complete revolution The speed of the particle would be the circumference of the circle of motion divided by the period Therefore, the period is defined as 2 r T v π
Image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Uniform Circular Motion, Force A force, , is associated with the centripetal acceleration The force is also directed toward the center of the circle Applying Newton’s Second Law along the radial direction gives 2 c v F ma m r = = r F r
Image of page 6
Example 1 - Conical Pendulum A 50 g ball is suspended from a string of length L = 80 cm. The ball revolves with constant speed in a horizontal circle of radius r. If θ = 12.2 0 ,what is the speed of the ball. The object is in equilibrium in the vertical direction and undergoes uniform circular motion in the horizontal direction ∑F y = 0 → T cos θ = mg ∑F x = T sin θ = m a c
Image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Example 2 - Motion in a Horizontal Circle A 0.25 kg mass is attached to the end of a cord 65 cm long. The ball is whirled in a horizontal circle. If the cord can withstand a maximum tension of 50 N, what is the maximum speed at which the ball can be whirled before the cord breaks? The centripetal force is supplied by the tension
Image of page 8
Example 3 - Horizontal (Flat) Curve A 1000-kg car moving on a flat, horizontal road negotiate a curve as shown. If the radius of the curve is 50 m and the coefficient of static friction between the tires and the road is 0.53, find the maximum speed the car can have and still make the turn successfully The force of static friction supplies the centripetal force: μ s mg =ma c a c = v 2 /R The maximum speed at which the car can negotiate the curve is Note, this does not depend on the mass of the car s v gr μ =
Image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 10
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern