A small block of mass M is placed halfway up on the inside of a frictionless, ci

Question

A small block of mass M is placed halfway up on the inside of a frictionless, circular loop of radius R (see the figure below). The size of the block is very small compared to the radius of the loop.

1)

Determine an expression for the minimum downward speed with which the block must be released in order to guarantee that it will make a full circle.

Answer is:

af=g=2fR

2f=gR

Ui+Ki=Uf+Kf

mgR+12m2i=mg2R+12mgR

i=sqrt(3gR)

Can someone please explain conceptually what is happening? I know that ac = v^2/R but how is this equal to g? And is "g" in this context referring to the gravitational constant 9.8? If so, how is this equal to the centripetal acceleration? I know this question has been asked and answered in terms of what the answer is, but none of those answers explain conceptually why/how they came to that conclusion. Thanks!

Explanation / Answer

To stay the block in circle, the forces act on it must be in equilibrium.

so Centripetal force = Weight

mv^2/R = mg

v^2/R = g

and finally ac = g

where  ac = centripetal acceleration

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