Circular MotionHard
Question
A small block slides with velocity 0.5
on the horizontal frictionless surface as shown in the Figure. The block leaves the surface at point C. The angle θ in the Figure is :


Options
A.cos-1 (4/9)
B.cos-1 (3/4)
C.cos-1 (1/2)
D.none of the above
Solution
Given vB = 0.5 
Assume block leave the contact at C, N = 0
= mg cos θ .... (1)
from energy conservation
mvB2 + mgr (1 − cos θ) =
mvC2 ........... (2)
from equation (1) and (2).
m
+ mg r (1 − cos θ) =
mg r cos θ
⇒ cos θ =
⇒ θ = cos-1
Ans.
Assume block leave the contact at C, N = 0
from energy conservation
from equation (1) and (2).
⇒ cos θ =
⇒ θ = cos-1
Create a free account to view solution
View Solution FreeMore Circular Motion Questions
A stone of mass of 16 kg is attached to a string 144 m long and is whirled in a horizontal smooth surface. The maximum t...A body of mass 5kg is moving in a circle of radius 1 m with an angular velocity of 2rad/ sec.Then the centripetal force ...For a particle in uniform circular motion the acceleration at a point P(R, θ ) on the circle of radius R is (here &...A particle is moving in a circle :...A particle moves in a circle of radius 5 cm with constant speed and time periof 0.2 π s. The accleration of the par...