Join now for JEE/NEET and also prepare for Boards Join now for JEE/NEET and also prepare for Boards. Register Now
Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
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¢ 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¢
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(Rsinθ−x)=Mx mRsinθ=(M+m)xmRsinθ=(M+m)xDifferentiate wrt time :mRωcosθ=(M+m)vmRωcosθ=(M+m)vwhere v=x˙v=x˙ and ω=θ˙ω=θ˙. We are given that M=4mM=4m soω=5v/Rcosa
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)=MxmRsin© = (m+M)x |||| becomes Rsin© = 5xDiff.both sides w.r.t time.Therefore, equation reduces to Rwcos©=5vd©/dt =w and dx/dt = v.Therefore, w=5v/Rcos© .
Dear student,Please find the solution to your problem. There are no external forces acting horizontally so the CM remains in the same position horizontallywhen hemisphere moves towards right by a small distance x with velocity v, body declines by an angle θ.Therefore, m(Rsinθ – x) = MxmRsinθ = (m + M)xNow, M = 4mTherefore, Rsinθ = 5xDiff. both sides w.r.t time.Rcosθ.dθ/dt = 5.dx/dtdθ/dt = w and dx/dt = vRcosθw = 5vTherefore, w = 5v/Rcosθ Thanks and regards,Kushagra
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -