A dead tree is initially stationary and there is a 30.0-degree angle between the tree and the vertical. The roots finally give way and the tree topples to the ground (rotating on an axis that is parallel to the level ground and intersecting the base of the tree perpendicularly through the base of the tree. The length of the tree is 30.0 m. How fast is the top of the tree moving (in m/s) just before it hits the ground? Model the tree as straight, rigid rod, and ignore friction and air resistance
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By the law of energy conservation:-
=>KE(final) = PE(initial)
=>1/2mv^2 = mgh
=>v = √2gh
=>v = √[2 x 9.8 x 15cos30*] {the C.G. will be at the center of the length}
=>v = √254.61
=>v = 15.96 m/s
By v = rω
=>ω = 15.96/15 = 1.06 rad/sec
Thus velocity of top (V) = rω = 30 x 1.06 = 31.91 m/s
=>KE(final) = PE(initial)
=>1/2mv^2 = mgh
=>v = √2gh
=>v = √[2 x 9.8 x 15cos30*] {the C.G. will be at the center of the length}
=>v = √254.61
=>v = 15.96 m/s
By v = rω
=>ω = 15.96/15 = 1.06 rad/sec
Thus velocity of top (V) = rω = 30 x 1.06 = 31.91 m/s
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Assume the tree falls at g and model the center of mass as being 15m up the tree
h = 15sin60 = 13m Assume mgh = 1/2 m*V^2 at impact
13mg = 1/2 mV^2 => 26g = V^2 => V = 16m/s BUT that is the middle of the tree, at twice that distance, V = 31.9m/s
h = 15sin60 = 13m Assume mgh = 1/2 m*V^2 at impact
13mg = 1/2 mV^2 => 26g = V^2 => V = 16m/s BUT that is the middle of the tree, at twice that distance, V = 31.9m/s