Motion of Object in a Vertical Circle
Consider an object of mass m tied to the end of a string and whirled in a vertical circle of radius r as shown in the figure. As it goes from lowest point A to highest point C its velocity goes on decreasing and becomes minimum at C and maximum at A. So, in vertical motion, the speed of the object does not remain constant.
Let vA, vC are the velocities of the object at A and C and TA, TC be the tension in the string at these points respectively. Let T be the tension on the string and v be the velocity of the object when the object is at point P such thatT - mg cosθ = mv2/r
or, T = (mv2/r) + mg cosθ …....... (i)
At the lowest point A, θ = 0. So, cos θ = 1. Then TA = (mvA2/ r) + mg .....… (ii)
At the highest point C, θ = 180. So, cosθ = -1. Then TC = (mvC2/ r) – mg .....… (iii)
At point B, θ = 90, so, cosθ = 0. Then TB = mvB2/ r ......… (iv)
From (ii), (iii) and (iv) it is clear that the tension on the string is maximum when the object moving with speed v is at lowest point A and is minimum when the object is at highest point C i.e.
Tmax = TA = (mv2/ r) + mg
Tmin = TC = (mv2/ r) – mg
If the tension at C is zero i.e. T = 0 then,
mg = mvC2/ r
or, vC = √ r. g
This is the maximum velocity as C to complete the circle called critical velocity. If the velocity of the object at the highest point C is less than this value, the string gets slack and the object will not complete the circle.
To calculate the minimum speed at A to complete the loop, we proceed as follows:
Total mechanical energy at A = Total mechanical energy at C
(P.E. + K.E.) at A = (P.E. + K.E.) at C
or, K.E. at A = (P.E. + K.E.) at C – P.E. at A
or, K.E. at A = (P.E. at C – P.E. at A) + K.E. at C
or, ½ mvA2 = mg (2r) + ½ mvC2
or, vA2 = 4gr + vC2
Since, vC2 = rg
vA2 = 4gr + gr
.: vA = √ 5gr
This equation gives the magnitude of the velocity at the lowest point with which a body can safely go around the vertical circle of radius r.
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