Yes, I had been making a mistake.
Consider the ring with the point mass as the system. The net force acting on it is 2mg-N at the time of release. Now, 2mg-N=2maCOM. From the centre of the ring, the only torque is Rmg acting on the point mass. Now, Rmg=Iα=2mR2α ie g=2Rα.
The centre of mass is always at the mid-point of the centre of the ring and the position of the point mass. So, aCOM=R2α.
From this, we can get the value of N as 32mg.
4 Answers
Soumyadeep Basu
·2013-09-17 10:53:48
Soumyadeep Basu
·2013-09-12 10:29:51
No, it will be 43mg. Is it correct now?
- Niraj kumar Jha No ,the ans is 3mg/2
Upvote·0· Reply ·2013-09-16 08:45:44
Soumyadip Kundu
·2014-02-05 07:49:27
Assume that a geosynchronous communications sattelite is in orbit at kolkata [latitude 28°]you are in kolkata and want to pick it signals. In what direction should you point the axis of the parabolic antena?