InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A spherical ball of mass m and radius r rolls without slipping on a rough concave surface of large radius R. It makes small oscillations about the lowest point. Find the time period. |
|
Answer» Correct Answer - B Let the angular velocity of the system about the point is suspension at any time be `omega` `Sve=(R-r)omega` `Again ve=romega` `[where omega_1=`rotatioN/Al velocity of the sphere] `1-v_c/r=((R-r)/r)omega` ………..1 By energy method total energy in SHM is constant. `So, `mg(R-r)(1-costgheta)+/2mv_c^2+1/2Iomega^2` =constant `:. mg(R-r))1-costheta)+1/2m(R-r)^2omega^2`=constant `:. mg(R-r)(1-costheta)+1/2m(R-r)^2omega^2+1/2mr^2((R-r)/r)omega^2=constant`:.mg(R-r)(1-costheta)+1/2m(R-r)^2omega^2+1/2mr^2((R-r)/r)omega^2` =constant `rarr g(R-r)(1-costheta)+(R-r)omega^I2[1/2+1/2]` =costant Taking derivative `g(R-or)sintheta(dtheta)/(dt)=7/10 (R-r)^22omega (domega)/(dt)` `rarr gsintheta=2xx(7/10)(R-r)alpha` `implies gsin theta=(7/5)(R-r)alpha` `alpha=(5gsintheta)/(7(R-r))` `:. alpha/theta=omega^2` `=(5g)/(7(R-r))=constant So the motion is SHM again `omega=sqrt((5g)/(7(R-r)))` `rarr T=2pi sqrt((7(R-r))/(5g))` |
|