Calc Forces on Hinge for Dynamically Loaded Stick

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SUMMARY

The discussion focuses on calculating the forces on a frictionless hinge for a uniform stick of length L and mass M, released from an angle theta_0. The key equation derived is the tangential force exerted on the stick by the hinge, expressed as F_t = Mg/4*sin(theta). Participants emphasize the importance of applying Newton's second law to determine the acceleration of the center of mass to find the reaction force accurately.

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  • Familiarity with torque and rotational motion concepts.
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  • Basic proficiency in trigonometric functions and their applications in physics.
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  • Study the derivation of forces in dynamic systems using Newton's second law.
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  • Learn about the center of mass and its significance in analyzing motion.
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Students studying physics, particularly those focusing on dynamics and rotational motion, as well as educators seeking to clarify concepts related to forces on hinges and dynamic loading scenarios.

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Homework Statement



In this problem we want to learn a little bit about what is sometimes called dynamical loading. Our simple system consists of a uniform stick of length L and mass M hinged at one end. We would like to calculate the forces on the (frictionless) hinge when the stick is released from rest at an angle theta_0 with respect to the vertical. You may find it useful to combine work and energy equations with torque (N II) equations.

3.2 Show that the tangential (tangent to the direction of motion, perpendicular to the stick) force exerted on the stick by the hinge is F_t = Mg/4*sin(theta).

Homework Equations



F=ma
I=1/3mr^2
torque=r x F=I*alpha

The Attempt at a Solution



F*d=I*alpha
(L/2)mgsin(theta)=(1/3)mL^2 * alpha

((3/2)mgsin(theta))/L = m*alpha

alpha = L*a

(3/2)mgsin(theta)=ma

Am I approaching this problem in the completely wrong way or am I missing something that factors into the tangental force? Any help is appreciated. Thanks!
 
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Hi Johnny0290 !:smile:

(have a theta: θ and an omega: ω and an alpha: α and a tau: τ and try using the X2 and X2 tags just above the Reply box :wink:)

Yes, that's fine until …
Johnny0290 said:
alpha = L*a

… but all that does is give you is the tangential acceleration of the end of the stick.

You need to find the acceleration of the centre of mass of the stick, so that you can apply good ol' Newton's second law to find the "missing" reaction force :wink:
 

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