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Force at the hinge

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Two masses each of m are attached at mid point B & end point C of massless rod AC which is hinged at A. It is released from horizontal position as shown. Find the force at hinge A when rod becomes vertical .

I tried to conserve energy to find velocity when it is vertical .

That is for mass at B

mgl/2 = (1/2)mv$^2$

From this I got centripetal force when vertical by mv$^2$/(l/2).

Same doing for mass at A .

I got as 2mg in both cases

So total force should be 6mg(including gravitational force .

But the answer given as 28mg/5

asked Dec 19, 2016 in Physics Problems by koolman (4,116 points)

1 Answer

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Best answer

This question is similar to the one posted 1 day ago by physicsapproval as "Force Exerted by Rod on the Hinge".

What is $v$ in your calculation? Particles B and C have different speeds. They are connected by a rigid rod so they have the same angular velocity $\omega$. If the velocity of B is $v$ then that of C is $2v$.

In this case the CM is $\frac34L$ from the axis so the decrease in PE is $(2m)g(\frac34L)$. The KE gained is $\frac12I\omega^2$ where $I=m((\frac{L}{2})^2+L^2)=\frac54mL^2$ is the moment of inertia of the rod.

The total decrease in PE must equal the total increase in KE :
$\frac12(\frac54mL^2)\omega^2=\frac32mgL$
$\omega^2=\frac{12g}{5L}$.

The centripetal force required to keep the rod moving in a circle is
$(\frac{L}{2}+L)m\omega^2=\frac32mL(\frac{12g}{5L})=\frac{18}{5}mg$.
The force on the hinge is the weight of the rod plus the centripetal force keeping it moving in a circle, ie $2mg+\frac{18}{5}mg=\frac{28}{5}mg$.

answered Dec 19, 2016 by sammy gerbil (26,096 points)
selected Dec 19, 2016 by koolman
Why can we not conserve energy for each particle separately ?
Because that could lead to B and C having different angular velocities when the rod is vertical. B and C are attached to a *rigid rod*, which forces them to move through the same angles about A and to have the same angular velocity and to keep the same distances from each other and from A.
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