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