Asked by leonard
A small object of mass m= 30 kg slides down a spherical dome of radius R=12 m without any friction. It starts off at the top (polar angle θ=0) at zero speed. Use g=10 m/s2.
What is the magnitude of the force (in Newton) exerted by the dome on the mass when it is at θ=30∘
At what angle θ0 does the sliding mass take off from the dome? answer in degrees (0∘≤θ0≤90∘
What is the magnitude of the force (in Newton) exerted by the dome on the mass when it is at θ=30∘
At what angle θ0 does the sliding mass take off from the dome? answer in degrees (0∘≤θ0≤90∘
Answers
Answered by
camilo
the angle is NOT 90... because it reache a speed when it can "scape" ... the angle is then 85.3
the force is (costheta-sintheta+tantheta)*300
the force is (costheta-sintheta+tantheta)*300
Answered by
Anonymous
wrong answer
Answered by
Anonymous
I Conservation of energy
K0+U0=K1+U1
mgR(1-cos(th))=mv^2/2
II Point N=0:
mgcos(th)=mv^2/R
K0+U0=K1+U1
mgR(1-cos(th))=mv^2/2
II Point N=0:
mgcos(th)=mv^2/R
Answered by
Anonymous
if you solve II for cos(th) you get v from I, then you get the angle from II
but what is wron with N(30)=mgcos(th)?
but what is wron with N(30)=mgcos(th)?
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