Swirling!

Classical Mechanics Level 3

A particle of mass m m , tied to one end of a massless inextensible string, is whirled around in a horizontal circle in the absence of gravity. The length of the string is gradually reduced, keeping the angular momentum about the center of rotation constant.

If the tension in the string is given as T = k r n T = k r^n , where k k is constant and r r is the instantaneous radius of the circle, then find the value of n n .

-1 -2 -3 -4

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Tapas Mazumdar by Tapas Mazumdar , 20, India

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1 solution

Tapas Mazumdar
Nov 29, 2016

At any instant when the radius of circle is r r , we have the following two results:

Angular momentum ( L L ) = m r 2 ω = m r^2 \omega and,

Tension in the string ( T T ) = = Centripetal force on the particle due to its rotation = m r ω 2 = m r {\omega}^2

Combining the above two results, we obtain:

T = L 2 m r 3 = L 2 m ⋅ r − 3 T = \dfrac{L^2}{m r^3} = \dfrac{L^2}{m} \cdot r^{-3}

Since L L and m m are constants, therefore on comparison, we get,

k = L 2 m k = \dfrac{L^2}{m} and n = − 3 n = \boxed{-3}

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The centripetal force is m ( r ω 2 − d 2 r d 2 t ) m(r{\omega}^2 - \dfrac {d^2r}{d^2t} )

Abdelhamid Saadi - 4 years, 6 months ago

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yes, but at any instant, the centripetal force (which causes circular motion) is m w^2 r. This is because d^2r/dt^2 is a linear force. Basically the terminology is different. Both the forces in your expression are central, but only one is centripetal

Rohan Joshi - 2 months, 3 weeks ago

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