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A small bead is fixed on a circular loop of radius R as shown in the figure below. The loop is rotating about YY? axis with constant angular acceleration ‘α’. The loop starts from rest, then, the bead is in circular motion, then acceleration of the bead at instant ‘t’ is_______.  
question image
Asked by albinjijoamayapra | 13 Sep, 2019, 15:52: PM
answered-by-expert Expert Answer
Figure shows the forces acting on the bead when it is at P.  T is the Tension force connecting the bead to the centre R.
 
If the tension force is resolved in horizontal and vertical component, then the horizontal component provides
the centripetal force required for circular motion, and the vertical component T cos30 displaces
the bead along the loop against its weight.
 
Hence ,  T sin30 = T/2 = mω2R   or   T = 2mω2R
 
acceleration = ( T cos30 - m g )/m  = ( 2ω2R - g )
Answered by Thiyagarajan K | 14 Sep, 2019, 14:56: PM

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