Rectangular Plate SHM Time Period

[Tanya Sharma]

There will be an unbalanced torque (counterclockwise) about the CM .Just as the plate rotates,the tension T₂ goes to zero , ie. String 2 becomes slack .

Right ?

 

[TSny]

Yes. There will be a "critical angle" θ_c such that T₂ = 0 when θ = θ_c. The net torque about the CM will still be zero at this angle. But once θ > θ_c, string 2 will have become slack and there will be a nonzero CCW torque about the CM due to T₁. Then the plate rotates and the motion is more complicated.

You can see that if the plate were narrow horizontally and long vertically, then θ_c would be very small.

 

[Tanya Sharma]

Thanks for the insight .

Does finding the center of the circle require a bit of guesswork on our part ?

I first thought of the blue dot as the center .Found the coordinates with ehild's guidance.Then manipulated the coordinates by shifting the origin from the blue dot.

Was there a more systematic way to think about the center of the arc ?

 

[ehild]

The blue dot was the origin of the coordinate system. with respect to it, the coordinates of the CM were [Lsinθ , -(Lcosθ+a/2)]. That is, the X coordinate is X=Lsinθ, the y coordinate is Y=-Lcosθ-a/2: Lsinθ=X, -Lcosθ=Y+a/2. Square both equations and add them:

X²+(Y+a/2)²=L².

You got the equation of the circle the CM moves along. The centre is (0;-a/2). The radius is L. What was the guesswork?

ehild

 

[TSny]

The radius of the arc of the CM must equal the radius of the arc of the point of attachment of one of the strings to the plate. Thus, the radius of the arc of the CM must be the same as the length of the string.

Recall from post #24:

If the plate doesn't rotate, then it must move in pure translation. Thus the trajectories of any two points of the plate are congruent.

 

[Tanya Sharma]

The blue dot was the origin of the coordinate system. with respect to it, the coordinates of the CM were [Lsinθ , -(Lcosθ+a/2)]. That is, the X coordinate is X=Lsinθ, the y coordinate is Y=-Lcosθ-a/2: Lsinθ=X, -Lcosθ=Y+a/2. Square both equations and add them:

X²+(Y+a/2)²=L².

You got the equation of the circle the CM moves along. The centre is (0;-a/2). The radius is L. What was the guesswork?

ehild

Nice and simple. Thanks!

I didn't find the center ,the way you did . When I looked at the coordinates ,it struck to me that if I could shift the origin to (0,-a/2) ,that will do the trick .

That is the guesswork I was talking about .

 

[Tanya Sharma]

The radius of the arc of the CM must equal the radius of the arc of the point of attachment of one of the strings to the plate. Thus, the radius of the arc of the CM must be the same as the length of the string.

Nice!

Thank you very very much TSny and ehild .You both have been simply wonderful

 

[ehild]

Intuition is very important, and you have that ability. But it should be checked and proved by systematic work, which might not be that nice.

I followed the method applied in high-school problems. 'Prove that something is a circle and find the centre and radius'.
TSny-s method is more intuitive and physical : The plate only translates - all its points move along the same curve , the point of any attachment with a string has to move along a circle of radius L, so does the CM.

ehild