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- Homework Statement
- A uniform rod of length L pivoted at its upper end hangs vertically. It is displaced through an angle of 60° and then released. Find the magnitude of the force acting on a particle of mass 'dm' at the tip of the rod when the rod makes an angle of 37° with the vertical.

- Relevant Equations
- Tangential acceleration of tip(dm), a= L𝛂

After solving using energy conservation, I found the angular velocity at 37° to be

Tension and the weight

To find the resultant force, I resolved the centripetal force and tangential force to find the centripetal force as

and tangential force as

This would leave me with a tangential acceleration of

But when I calculated it using torque, I found the tangential acceleration to be

What am I doing wrong while resolving the forces?

I would also like to know what would be the forces provided by the part of rod that is above 'dm'?

**omega=2.97/(L)^½**Tension and the weight

**(dm)g**are the two forces acting on the tip*dm*To find the resultant force, I resolved the centripetal force and tangential force to find the centripetal force as

**F= (dm)(L)(omega)²**and tangential force as

**(dm)(g)sin37°**.This would leave me with a tangential acceleration of

**5.9 m/s²**But when I calculated it using torque, I found the tangential acceleration to be

**8.82m/s²**and that is the solution given in my bookWhat am I doing wrong while resolving the forces?

I would also like to know what would be the forces provided by the part of rod that is above 'dm'?