Objects moving in vertical circle with tension as the centripetal force
Example: A mass of 2kg tied to a string of 1m length is rotating in a vertical circle with a uniform speed of 4ms−1.Find the position of mass when tension in the string will be 52N.[Take g=10ms−2]
Solution: T−mgsinθ=Rmv2 (θ is the angle from the horizontal line Thus we get: 52−2×10×sinθ=12×16 Thus sinθ=90o Thus the tension of 52 N would occur at the bottom most point.
Objects in vertical circle with normal reaction as centripetal force
Example: A block is freely sliding down from a vertical height 4m on a smooth inclined plane. The block reaches bottom of inclined plane then it describes vertical circle of radius 1m along smooth track. What is the ratio of normal reactions on the block while it is crossing lowest point and highest point of vertical circle is:
Solution: Applying the law of conservation of energy at point A and B, we have mgh=21mv12⇒v1=2gh⇒v1=8g Now, net force is towards the center is the centripetal force, we have at point B rmv12=N1−mg ⇒N1=rmv12+mg=1m(8g)+mg=9mg ..........(I) applying the law of conservation of energy at point B and C, we have 21mv12=mg(2r)+21mv22⇒21m(8g)=mg(2)+21mv22⇒v2=4g Now, net force is towards the center is the centripetal force, we have at point C rmv22=N2+mg ⇒N2=rmv22−mg=1m(4g)−mg=3mg ..........(II) So the ratio of normal reactions on the block while it is crossing lowest point, highest point of vertical circle is: N2N1=3mg9mg=3:1 ..from (I) and (II)