Find Magnitude of Normal Reaction from Hinge after Release of Rod

[Mandeep Deka]

Homework Statement

A rod (length 'l'; mass 'm') is hinged to a vertical wall and is released when it makes an angle @ with the wall. Find the magnitude of normal reaction exerted by the hinge just after the release of the rod?

Homework Equations

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The Attempt at a Solution

I calculated the angular acceleration just after the release of the rod about the hinge point. Now this angular acceleration shall be the same as the angular acceleration of the hinge point about the center of mass of the rod. But if i solve this equation, i get (mgsin@/4) as the answer which is not correct!
Where is the problem in the solution?

Please help!

 

[anathema]

Thank you for your question. In order to accurately calculate the magnitude of the normal reaction exerted by the hinge just after the release of the rod, we need to consider the forces acting on the rod.

Firstly, we have the weight of the rod acting downwards, which is equal to mg. We also have the normal reaction force exerted by the hinge, which is equal in magnitude but opposite in direction to the weight of the rod. Additionally, we have the torque due to the weight of the rod acting about the hinge point, which is given by mglsin@.

Since the rod is released from rest, we can use the equation τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration. We can rewrite this equation as mglsin@ = Iα.

Next, we need to find the moment of inertia of the rod about the hinge point. This can be calculated using the formula I = (1/3)ml^2, where m is the mass of the rod and l is its length.

Substituting this into our equation, we get mglsin@ = (1/3)ml^2α. Solving for α, we get α = 3gsin@/l.

Now, we can use the equation F = ma to find the normal reaction force exerted by the hinge. Since the rod is in rotational motion, we need to use the rotational version of this equation, which is τ = Iα. This can be rewritten as F = Iα/l.

Substituting our values, we get F = (1/3)ml^2(3gsin@/l)/l = mgsin@.

Therefore, the magnitude of the normal reaction force exerted by the hinge just after the release of the rod is equal to mgsin@, which is the correct answer.

I hope this helps to clarify your doubts. Please let me know if you have any further questions.