A smooth circular hoop with a radius of 0.600 m is placed flat on the floor. A 0

Question

A smooth circular hoop with a radius of 0.600 m is placed flat on the floor. A 0.500-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 4.50 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. J (b) What is the total number of revolutions the particle makes before stopping? Assume the friction force remains constant during the entire motion. 2.0 rev

Explanation / Answer

here,

r = 0.6 m

initial speed , w0 = 8.5 /0.6 = 14.17 rad/s

final angular speed , w = 4.5 /0.6 = 7.5 rad/s

theta = 2 pi rad

let the angular accelration be alpha

using third equation of motion

w^2 - w0^2 =2 * alpha * theta

7.5^2 - 14.17^2 = 2 * alpha * 2 pi

alpha = - 11.5 rad/s^2

let the total angle travelled before stopping be theta'

0 - w0^2 = 2 * alpha * theta1

- 14.17^2 = - 2 * 11.5 * theta1

theta1= 8.72 rad

the total number of revolutions , n = theta1 /2pi = 1.39 rev

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