Asked by Anonymous
A small object of mass m= 90 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. (See figure)
(a) What is the magnitude of the force (in Newtons) exerted by the dome on the mass when it is at the top, at θ=0∘?
N(θ=0∘)=
(b) What is the magnitude of the force (in Newton) exerted by the dome on the mass when it is at θ=30∘?
N(θ=30∘)=
(c) At what angle θ0 does the sliding mass take off from the dome? answer in degrees (0∘≤θ0≤90∘; )
θ0=
(a) What is the magnitude of the force (in Newtons) exerted by the dome on the mass when it is at the top, at θ=0∘?
N(θ=0∘)=
(b) What is the magnitude of the force (in Newton) exerted by the dome on the mass when it is at θ=30∘?
N(θ=30∘)=
(c) At what angle θ0 does the sliding mass take off from the dome? answer in degrees (0∘≤θ0≤90∘; )
θ0=
Answers
Answered by
Anonymous
a) 300N
b) 230.38
c) Not sure
b) 230.38
c) Not sure
Answered by
dude
can u
pls provide the formulae?? please..!!
pls provide the formulae?? please..!!
Answered by
tiklam
a: mg (90*10=900)
b: No sure yet
c: Conservation of energy gives:
m*g*R(1-cos(theta))=0.5*m*v^2
at fly off there is no force on the dome. so m*g*R*cos(theta)=(mv^2)/R
Use these equations to eliminate v^2
you will get Cos-1(2/3)
Theta= 48.10 Degrees
b: No sure yet
c: Conservation of energy gives:
m*g*R(1-cos(theta))=0.5*m*v^2
at fly off there is no force on the dome. so m*g*R*cos(theta)=(mv^2)/R
Use these equations to eliminate v^2
you will get Cos-1(2/3)
Theta= 48.10 Degrees
Answered by
Ramana
doesnot be theta= 81.7872degree?
I have R=12m
I have R=12m
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