Question

A smooth circular hoop with a radius of 0.400 m is placed flat on the floor. A 0.325-kg particle slides around the inside edge of the hoop. The particle is given an initial speed of 8.50 m/s. After one revolution, its speed has dropped to 5.50 m/s because of friction with the floor.
(a) Find the energy transformed from mechanical to internal in the particle "hoop" floor system as a result of friction in one revolution.
(b) What is the total number of revolutions the particle makes before stopping? Assume the friction force remains constant during the entire motion.

Answer 1

Answer:

a)  W = - 6.825 J,  b) θ = 1.72 revolution

Explanation:

a) In this exercise the work of the friction force is negative and is equal to the variation of the kinetic energy of the particle

         W = ΔK

         W = K_f - K₀

          W = ½ m v_f² - ½ m v₀²

         W = ½ 0.325 (5.5² - 8.5²)

         W = - 6.825 J

b) find us the coefficient of friction

Let's use Newton's second law

            fr = μ N

y-axis (vertical)   N-W = 0

            fr = μ W

work is defined by

             W = F d

the distance traveled in a revolution is

             d₀ = 2π r

             W = μ mg d₀ = -6.825

            μ = $\frac{ -6.825}{d_o \ mg}$

               

The total work as the object stops the final velocity is zero v_f = 0

         W = 0 - ½ m v₀²

          W = - ½ 0.325 8.5²

          W = - 11.74 J

           μ mg d = -11.74

           

we subtitle the friction coefficient value

           ( $\frac{-6.8525 }{d_o mg}$) m g d = -11.74

               6.825  $\frac{d}{d_o}$ = 11.74

               d = 11.74/6.825  d₀

               d = 1.7201  2π 0.400

               d = 4.32 m

this is the total distance traveled, the distance and the angle are related

              θ = d / r

              θ = 4.32 / 0.40

              θ = 10.808 rad

we reduce to revolutions

              θ = 10.808 rad (1rev / 2π rad)

              θ = 1.72 revolution