Answer:
Explanation:
a)
[tex]KE_i + PE_i = KE_f + PE_f \\\\0 + mgh = 0.5 mv^2 + 0.5Iw^2\\\\I = mr^2 \\\\w= v/r \\\\mgh = 0.5 mv^2 + 0.5 mv^2 \\\\mgh = mv^2 \\\\9.8 x 7.58= v^2 \\\\v= 8.62 m/s[/tex]
b)
[tex]mgh= 0.5(m+\frac{mr^2}{2r^2})\times v^2 \\\\gh = \frac{3}{4} v^2 \\\\v= sqrt(\frac{4gh}{3}) m/s\\\\v= sqrt(\frac{4\times 9.8\times 7.58}{3}) \\\\v = 9.95 m/s[/tex]
(a)The final velocity of a hoop that rolls without slipping down a 7.58 m high hill, starting from the rest [tex]V_{F} = 8.62\dfrac{m}{s}[/tex]
(b) The final velocity of a disk of the same mass and radius as the hoop rolled down the hill [tex]V_{F} = 9.95\dfrac{m}{s}[/tex]
Given that
Height of the hill h= 7.58m
Acceleration due to gravity [tex]g=9.81\dfrac{m}{s}[/tex]
Now from conservation of energy, the total energy will remain constant
[tex]KE_{i} +PE_{i}= KE_{f}+PE_{f}[/tex]
[tex]0+mgh=0.5mV^{2}+0.5Iw^{2}[/tex]
[tex]I=mr^{2}\\ I=\frac{V}{r}[/tex]
[tex]mgh=0.5mv^{2}+ 0.5mv^{2}=mv^2[/tex]
[tex]mgh=mv^2[/tex]
[tex]9.81\times 7.58= v^2[/tex]
[tex]V_{F}=8.62\frac{m}{s^2}[/tex]
(B) For the second case
[tex]mgh=0.5M+\dfrac{mr^2}{2r^2} V^2[/tex]
[tex]gh=\dfrac{3}{4} V^2[/tex]
[tex]V=\sqrt{\dfrac{4gh}{3} }[/tex]
[tex]V=\sqrt{\dfrac{4\times9.81\times7.58}{3} }[/tex]
[tex]V_{F} =9.95\frac{m}{s}[/tex]
Thus,
(a)The final velocity of a hoop that rolls without slipping down a 7.58 m high hill, starting from the rest [tex]V_{F} = 8.62\dfrac{m}{s}[/tex]
(b) The final velocity of a disk of the same mass and radius as the hoop rolled down the hill [tex]V_{F} = 9.95\dfrac{m}{s}[/tex]
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