Question

Please explain the steps to get answer B)

24. A hoop of mass M and R=0.600 m is released from rest and rolls without slipping down an incline that is at an angle of 60° above the horizontal. The moment of inertia of a hoop about its center of mass is I ☆ = MRC. When the hoop has traveled a distance S= 4.0 m down the incline, the magnitude of its angular velocity about its center is A) 7.38 rad/s. B) 9.71 rad/s. C) 11.2 rad/s. D) 10.4 rad/s. E) 14.8 rad/s.

Answer 1

here,

the radius of hoop , R = 0.6 m

s = 4 m , theta = 60 degree

the height dropped , h = s * sin(theta)

h = 4 * sin(60) m = 3.46 m

let the final angular velocity be w

using conservation of energy

the kinetic energy gained = potential energy lost

rotational kinetic energy + translational kinetic energy = potential energy lost

0.5 * I * w^2 + 0.5 * M * v^2 = M * g * h

0.5 * (M * r^2) * w^2 + 0.5 * M * (w * r)^2 = M * g * h

w^2 * r^2 = g * h

w^2 * 0.6^2 = 9.81 * 3.46

solving for w

w = 9.71 rad/s

the magnitude of angular velocity is B) 9.71 rad/s