#50 second hill is cireutal wil resistance. What must be the height h of the fir

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#50

second hill is cireutal wil resistance. What must be the height h of the first hill so t loses contact with the snow at the crest of the second hill? in on (abo don tota ati * 57 gi **50. A person starts from rest at the top of a large frictionless spherical surface, and slides into the water below (see the draw- ing). At what angle does the person leave the surface? (Hint: When the person leaves the surface, the normal force is zero.) Section 6.6 Nonconservative Forces and the Work-Energy Theorem 1m A prgjectile of mass 0.750 kg is shot straight up with an initial

Explanation / Answer

Normal force is N, centrifugal force F = mgsin(t)- N. At the given point, N= 0. Also, it is circular motion so F= mv^2/r. So, rgsin(t)= v^2

from geometry , sin(t) = h/R where h is its height from floor when it leaves the surface.

from conversation of energy, mgr=mv^2/2+mgh =>h=r(1-0.5sin(t))= r(1-0.5h/r). So, h= r-0.5h , so h=2r/3

So, sin(t)= 2r/3 / r = 2/3

So , t= sin^-1(2/3)= 41.8°

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