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Normal reaction in circular motion

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A bead of mass m is tied at one end of a spring of spring constant mg/R and unstretched length R/2 and other end to fixed point O. The smooth semicircular wire frame is fixed in vertical plane. Find the normal reaction between bead and wire just before it reaches the lowest point.

I tried as
By energy conservation
PE of spring(final - initial) + PE due to gravity = change in KE
(1/2)(mg/R)R$^2$ + 2mgR = (1/2)mv$^2$
From this I got cetripetal force as 5mg towards the centre,O.
And gravitational force ,mg at B downwards .
So net force is 4mg that is normal reaction

But the answer given as 6mg

asked Jan 27, 2017 in Physics Problems by koolman (4,116 points)
edited Jan 27, 2017 by koolman
Please can you explain your equation for energy conservation? I don't understand it. And how did you get centripetal force $5mgR$? This is not a force. $5mg$ is a force, $5mgR$ is not.
Sorry for the mistake , I have edited
How can you post questions like bullets from machine gun whereas I have to wait for approval ?
Hi Moto at 200 reputation you will no longer require approval.

1 Answer

2 votes
 
Best answer

You have subtracted $mg$ from $5mg$ but you should have added.

The centre of rotation is above B so the net upwards force on the bead is
$N-mg=\frac{mv^2}{R}=5mg$
where $N$ is the normal reaction $N$ acting on the bead. Hence
$N=6mg$.

answered Jan 27, 2017 by sammy gerbil (26,096 points)
selected Jan 28, 2017 by koolman
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