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Question

A bead of mass m is attached to one end of a spring of natural length R and spring constant k=mgR. The other end of the spring is fixed at point A on a smooth vertical ring of radius R as shown in the figure. The tangential acceleration of the bead just after it is released will be:

A
13.4g cm/s2
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B
0.134g cm/s2
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C
13.4g m/s2
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D
Can't be determined
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Solution

The correct option is A 13.4g cm/s2
From the geometry shown in figure, it is evident that ABR and there is an extension in the spring , let's say x.


x= length of AB - natural length of spring
x=(Rcos30+Rcos30)R
=2Rcos30R=(31)R
Now, spring force:
(F)=kx=mgR(31)R=(31)mg

From FBD:

Applying equation of circular motion along tangential direction:
mgsin30Fsin30=m×at
mat=mg2F2
mat=mg2(3 1) mg2
mat=2 mg3 mg2
at= (23)g2

|at|=0.134g m/s2=13.4g cm/s2

If actual value of at comes negative, then it represents that actual direction of at is opposite of what we assumed in FBD.

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