Calculating the tesion in a string holding a 1 meter stick

[saurabheights]

A meterstick (L = 1 m) has a mass of m = 0.13 kg. Initially it hangs from two short strings: one at the 25 cm mark and one at the 75 cm mark.



What is the tension in the left string right after the right string is cut?



Calculating the free body diagram where we put the acceleration of center of mass of the plank, mg and T.

 

[Doc Al]

Hint: Consider torques about the point where the left string attaches to the stick.

 

[saurabheights]

Yeah did that. I first calculated torque about the left string using integral calculas over a thin piece of stick at distance x. The total torque calculated was MGL/4. Then calculated Inertia. It came out to be 7ML^2/48. Then, I calculated angular acceleration as = 7g/12L.

Finally, I calculated tension using free body diagram → T-mg+ ∫(dm * α * x) = 0
→T-mg+∫(M/L*α*x.dx) = 0
where x went from -0.25 to +0.75.

T = mg-∫(ML*dm/L*x.dx)

Is this the correct solution?

 

[Doc Al]

Yeah did that. I first calculated torque about the left string using integral calculas over a thin piece of stick at distance x. The total torque calculated was MGL/4. Then calculated Inertia. It came out to be 7ML^2/48.

Good.

Then, I calculated angular acceleration as = 7g/12L.

Redo that calculation.

Finally, I calculated tension using free body diagram → T-mg+ ∫(dm * α * x) = 0

Not sure what you are doing here or what that last term represents.

Just use: ƩF = ma