# IIT JEE Main / Advanced Physics 2018 : PHYSICSMYND Brainwave(r) > Post.1

Physics is a scientific course , where excellence comes from imbibing the principles to a level where you become a master magician [ of sort ] dabbling in theoretical potions ( ! ) to bring out unique perspectives and intuitive problem solving skills. Well , to get to the point, this series of ours deals with physics concepts / problems , where your skills will be put to the maximum test . They are of a level usually asked in competitions like the Physics Olympiad , and is meant to sharpen your problem solving skills for the IIT JEE Main / Advanced Physics paper.

The case of the MERRY – GO-ROUND A merry-go-round accelerates from rest with a constant angular acceleration of 0.02 revolutions /second/second.A passenger sitting on a chair 6 meters from the axis of rotation holds a 2 kg weight. Find out the magnitude and direction of the force he /she must exert to hold the weight 5 seconds after the merry-go-round starts to rotate. Also specify the direction with respect to the radius of the chair  which the passenger is using ? In this case, we have to consider two co-ordinate frames, both having the same origins, but one , say L ,which is fixed and the other R, which rotates with an angular velocity ω . Consider a vector A , whose time derivatives in the above mentioned two frames are related by –

( dA/dt )L =  (dA/dt )R + ω x A  , now for a point of radius vector r from the origin , it’s –

( dr/dt )L =  (dr/dt )R + ω x r , and  ( d²r/dt² )L =  (d/dt )L. (dR/dt )R + ω x ( dr/dt )L + (dω/dt )Lx r

As,   (d/dt )L (dr/dt )R = ( d²r/dt² )R + ω x  (dR/dt )R and

ω x (dr/dt )L =    ω  x (dr/dt )R +   ω x ( ω x r  )            now,

(d²r/dt² )R =  a’   ,  ( dr/dt )R = v’   and   (dω/dt )L = ω’    ,  hence , we have –

(d²r/dt² )L =  a’  +  2ω x v’  + ω x(ω x r )  + ω’ x r . Now , considering the rotating frame which is attached to the merry-go-round, the equation for the weight’s motion , whose F = m  (d²θ/dt² )L , now gives  –

ma’   =      F + mω²r      =      mω’ x r – 2mω  x v’

note that , here ω is perpendicular to r . Now. with respect to the merry-go-round , the weight is stationary and hence , a’ = 0 and v’ = o , therefore the above equation becomes , F  = – mω²r  + mω’  x r

Considering the rotating frame R , we can take an z axis along the axis of rotation and the z axis from the center towards the passenger’s chair-

then,   ω’  = ω’ k   and  r  = ri ,  Now , the force acting on the weight is the resultant of the holding force f  exerted by the passenger + the gravity of the earth ,     hence , F = f – mgk   and so  f =  – mω²ri  +  mω’rj  +  mgk   , now , we have-

ω’  = 0.02  x   2π rad/s²  ,  ω = 5 ω’  , m = 2 kg  and r = 6m  ,   and f = -4.74i  + 1.51j + 19.6 kN   = 20.2 N