Question

A hoop (cylindrical shell) is rolling without slipping along a horizontal surface with a forward (translational) speed of 6.51 m/s when it starts up a ramp (still rolling without slipping) that makes an angle of 25.0° with the horizontal. What is the speed of the hoop after it has rolled 3.48 m up as measured along the surface of the ramp?

a) 5.29 m/s

b) 6.50 m/s

c) 5.97 m/s

d) 3.91 m/s

e) 2.70 m/s

Answer 1

here,

the initial velocity of hoop , u = 6.51 m/s

theta = 25 degree

s = 3.48 m

let the final velocity be v

using conservation of energy

KEi + Pei = KEf + Pef

(0.5 * I * w0^2 + 0.5 * m * u^2) = (0.5 * I * w^2 + 0.5 * m * v^2) + m * g * (s * sin(theta))

(0.5 * m * r^2 * (u/r)^2 + 0.5 * m * u^2) = (0.5 * m * r^2 * (v/r)^2 + 0.5 * m * v^2) + m * g * (s * sin(theta))

(u^2) = (v^2) + g * (s * sin(theta))

6.51^2 = v^2 + 9.81 * 3.48 * sin(25)

solving for v

v = 5.29 m/s

the final velocity of hoop is A) 5.29 m/s