Question

Calculate the final speed of a wheel that starts from rest and rolls down a 12.00-m-high incline. The wheel can be approximated to a 0.9 kg (M_H) hoop spinning about its central axis with a radius (R_H) of 0.35m and three 0.2 kg (M_R) rods of 0.35 m (L_R) length rotating from the axis. Assume the total mass of the wheel is comprised of just a hoop and 3 rods and assume the radius of rotation is 0.35m.

Answer 1

here,

height of incline , h = 12 m

the mass of hoop , mH = 0.9 kg

the radius of hoop , r = 0.35 m

mass of each rod , mR = 0.2 kg

the moment of inertia of wheel , I = mH * r^2 + 3 * mR * r^2 /3

let the final velocity be v

using conservation of energy

potential energy lost = kinetic energy gained

((mH + 3 mR) * g * H ) = 0.5 * (mH + 3 mR) * v^2 + 0.5 * I * w^2

((mH + 3 mR) * g * H ) = 0.5 * (mH + 3 mR) * v^2 + 0.5 * (mH * r^2 + 3 * mR * r^2 /3) * (v/r)^2

((0.9 + 3* 0.2) * 9.81 * 12 ) = 0.5 * (0.9 + 3 * 0.2) * v^2 + 0.5 * (0.9 * 0.35^2 + 3 * 0.2 * 0.35^2 /3) * (v/0.35)^2

solving for v

v = 11.65 m/s

the final speed of the wheel is 11.65 m/s