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Quick question, I'm a bit confused by this problem:

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A mass $m$ attached to a string is moving horizontally with constant velocity: Clearly angular momentum with respect to $M$ is conserved. Is angular momentum with respect to point A conserved as well in this setup? I'd say it isn't because due to the momentum pointing tangentially along the circular path of the mass and $l \times p$ changes direction whilst the mass $m$ is moving. So what's the torque acting on the mass? Is it the constraint force (cross $r$) of the string?

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You have correctly pointed out that angular momentum about A (red arrow) will change. Forces on mass m are Tension and Gravity. Now if you consider torque about point A then torque due to tension becomes zero because tension force is parallel to l but torque due to gravity is non zero and that torque provides the change in angular momentum.

$\vec{\tau}= \vec{l} \times m\vec{g} $

Torque due to gravity about point A is along the tangent of the circle (green arrow). So, torque will only induce a change of angular momentum only along the tangential direction, so vertical component of angular momentum doesn't change, only horizontal component changes. The left side of the figure shows this clearly.

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Let's call $\vec h=AM$, such that $\vec l=\vec h+\vec r$. Then the torque $$\vec \tau=\frac{d}{dt}(\vec l\times \vec p)=\frac{d}{dt}((\vec h+\vec r)\times \vec p)=\vec h\times \frac{d}{dt}\vec p+ \frac{d}{dt}(\vec r\times \vec p)$$ The last term is zero, since the angular momentum with respect to $M$ is conserved. I've also used the fact that $\vec h$ is a constant. Then, since $\vec p$ is always in the horizontal plane, $d\vec p/dt$ is also in the horizontal plane, so the force you are interested in is the resultant of the tension in the string and gravity.

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  • $\begingroup$ The angular velocity with respect to A changes in direction, but the magnitude is constant $\endgroup$ – Andrei Dec 5 '17 at 18:28
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It is clear that the motion of the mass is in horizontal plane & not having a single effect in vertical plane (i.e. no acceleration in vertical plane) If we resolve the tension (T) on the mass due to string.. We get, T sin (alpha) as the centrifugal force ( direction :- radially inwards, in plane of the circle). T cos (alpha) in vertical plane (direction:- upwards ). And we have mg as the gravitational force acting in opposite directions to Tcos(alpha) . Now, As there is no acceleration in vertical plane, the magnitude of Tcos(alpha) will be equal to mag. of mg..

Considering torques of all the present forces about point A on the axis AM, Torque due to centripetal force Tsin(alpha) will be ZERO. In vertical plane, magnitudes of torques due to Tcos(alpha) & mg will be equal to Tcos(alpha)r & mgr respectively.. Tcos(alpha) = mg.. so both torques will be of equal magnitude and will be in opposite directions cancelling each other...

So, No torque acts due to forces in vertical plane.

Hence, No torque is acting about point A on axis AM..... .. angular momentum is conserved...

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  • $\begingroup$ If we consider an axis passing through point A & parallel to plane of horizontal circle, the only torque acting will be due to centripetal force (Tsin(alpha)) will be present & will vary in it's directions.. So will the angular momentum be conserved about this axis???? $\endgroup$ – user177698 Dec 5 '17 at 19:24

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