A string is wrapped several times times around the rim of a

A string is wrapped several times times around the rim of a small hoop with radius 8.00cm and mass 0.180kg.the free end is held in place and the loop is released from rest.after the hoop has descended 75.0cm,calculate:
a) angular speed of rotating hoop?
b) speed of its center?
坏习惯成自然 1年前 已收到3个回答 举报
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天啊真的累gg我了 幼苗

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Moment of inertia (I) for a hoop,I = mr²
PE lost (mgh) = rotational KE gained (½Iω²) + linear KE gained (½mv²)
mgh = ½(mr²)ω² + ½mv²
2gh = r²ω² + v²
as v = rω ,v² = r²ω²
2gh = 2.r²ω²
gh = r²ω²
ω = √(gh) / r =√(9.80 x 0.75m) / 0.08m
ω = 33.90 rad/s

1年前

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王小野 幼苗

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翻一下吧，看不懂

1年前

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小木an 幼苗

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拿售后检修吧

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