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Mayhem
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- Homework Statement
- A bar of mass m is attached to a pivot point and creates an angle φ with horizontal. It moves counter-clockwise from the initial angle, φ_0, to the final angle, φ_1. What is the total work done by the bar?
- Relevant Equations
- F=mg*cos(φ)
This isn't really a proper homework question, but something I wondered about myself. To simplify things, we say that pivot point is in the origin of a Cartesian coordinate system, and the angles are constrained to the first quadrant.
We see that the weight of the barbell is given as $$F = mgcos(\phi)$$. Since in translatoral mechanics, we define work as $$W = \int_{x_0}^{x_1} ma dx$$, would it make sense in rotational mechanics to define work as $$W = \int_{\phi_0}^{\phi_1} ma\cos{\phi} d\phi$$ ?
In that case, I get the general solution of the work being done as
$$W = mg(\sin{\phi_1}-\sin{\phi_0})$$
We see that the weight of the barbell is given as $$F = mgcos(\phi)$$. Since in translatoral mechanics, we define work as $$W = \int_{x_0}^{x_1} ma dx$$, would it make sense in rotational mechanics to define work as $$W = \int_{\phi_0}^{\phi_1} ma\cos{\phi} d\phi$$ ?
In that case, I get the general solution of the work being done as
$$W = mg(\sin{\phi_1}-\sin{\phi_0})$$
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