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Grade 12Mechanics

A hemisphere of radius R and mass 4m is free to slide with it`s base on a smooth horizontal table. A particle of mass m is placed on the top of the hemisphere. The angular velocity of the particle relative to hemisphere at an angular displacement ¢ when velocity of hemisphere is v is A) 5v/Rcos¢ B) 2v/Rcos¢ C) 3v/Rcos¢ D) 6v/Rcos¢

abhishek sabnis

9 Years agoGrade 12

3 Answers

lj

8 Years ago

When the hemisphere has moved right by distance xx the particle has slid down through angle θθ. There are no external forces acting horizontally so the CM remains in the same position horizontally :m(Rsinθ−x)=Mxm(Rsin⁡θ−x)=Mx mRsinθ=(M+m)xmRsin⁡θ=(M+m)xDifferentiate wrt time :mRωcosθ=(M+m)vmRωcos⁡θ=(M+m)vwhere v=x˙v=x˙ and ω=θ˙ω=θ˙. We are given that M=4mM=4m soω=5v/Rcosa

VISHAL Yadav

7 Years ago

There is external force acting therefore CM remains in the same position horizontally. Now when hemisphere suppose moves towards right by a small distance x with velocity v, body declines by angle © theta. Therefore, m(Rsin©-x)=Mx

mRsin© = (m+M)x |||| becomes Rsin© = 5x

Diff.both sides w.r.t time.

Therefore, equation reduces to Rwcos©=5v

d©/dt =w and dx/dt = v.

Therefore, w=5v/Rcos© .

Kushagra Madhukar

5 Years ago

Dear student,

Please find the solution to your problem.

There are no external forces acting horizontally so the CM remains in the same position horizontally

when hemisphere moves towards right by a small distance x with velocity v, body declines by an angle θ.

Therefore, m(Rsinθ – x) = Mx

mRsinθ = (m + M)x

Now, M = 4m

Therefore, Rsinθ = 5x

Diff. both sides w.r.t time.

Rcosθ.dθ/dt = 5.dx/dt

dθ/dt = w and dx/dt = v

Rcosθw = 5v

Therefore, w = 5v/Rcosθ

Thanks and regards,

Kushagra

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