Answer:
a) 70.5117 J
b) 47.01 J
c) 7.05 m
d) 4.7 m
Explanation:
a)The rotating wheel has both translational K.E. (linear kinetic energy) and rotational kinetic energy.
Total K.E. of rotating wheel = [tex]\frac{1}{2} mu^{2} + \frac{1}{2}I[/tex]ω²
=[tex]\frac{1}{2}[/tex](mu² + [tex]\frac{1}{2}[/tex]mr²×[u/r]²)
where ω = angular velocity and r = radius of the wheel
we get K.E. = [tex]\frac{3mu^{2} }{4}[/tex]
= 0.75×2.71×(5.89)²
= 70.5117 J
b) for sliding wheel, it has just translational K.E.
Total K.E. of sliding wheel = [tex]\frac{1}{2} mu^{2}
= 0.5×2.71×(5.89)² = 47.01 J
c) derivation of the following equation is in the attachment,
if h = height climbed by the rotating wheel
h = [tex]\frac{3mu^{2} }{4g}[/tex]
= 0.75×2.71×(5.89)²/10
= 7.05 m
d) derivation of the following equation is in the attachment,
if H = height climbed by the sliding wheel
H = [tex]\frac{mu^{2} }{2g}[/tex]
= 0.5×2.71×(5.89)²/10
= 4.7 m
