A solid uniform disk of diameter 3.20 m and mass 42 kg rolls without slipping to the) bottom of a hill, starting from rest. If the angular speed of the disk is 4.27 rad/s at the bottom, how high did it start on the hill? A) 3.57mB) 4.28 mC) 3.14 mD) 2.68 m

(A) = 3.57 m

Explanation:

from the question we are given the following:

diameter (d) = 3.2 m

mass (m) = = 42 kg

angular speed (ω) = 4.27 rad/s

from the conservation of energy

mgh = 0.5 mv^{2} + 0.5Iω^{2} ... equation 1

where

Inertia (I) = 0.5mr^{2}

ω = / frac{v}{r}

equation 1 now becomes

mgh = 0.5 mv^{2} + 0.5 (0.5mr^{2}) (/frac{v}{r}) ^{2}

gh = 0.5 v^{2} + 0.5 (0.5) (v) ^{2}

4gh = 2v^{2} + v^{2}

h = 3v^{2} : 4 g ... equation 2

from ω = / frac{v}{r}

v = ωr = 4.27 x (3.2 : 2)

v = 6.8 m/s

now substituting the value of v into equation 2

h = 3v^{2} : 4 g

h = 3 x (6.8) ^{2} : (4 x 9.8)

h = 3.57 m