Rectangular Plate SHM Time Period

[Tanya Sharma]

Homework Statement

A rectangular plate of sides ‘a’ and ‘b’ is suspended from a ceiling by two parallel strings of length L each. The separation between the strings is ‘d’. The plate is displaced slightly in its plane keeping the strings tight. Find the time period of SHM.

Homework Equations

The Attempt at a Solution

I have drawn two rough sketches depicting the orientation of the plate while in motion.Not sure which one correctly describes the situation .In the first picture the side ‘b’ always remain horizontal while in 2nd picture the side ‘b’ makes an angle θ with the horizontal.θ is the angle which strings make with the vertical.

I think the 1st picture is correct . In that case the plate is not rotating about the CM and the plate can be simply replaced by its CM .

The CM is at a distance L+a/2 from the ceiling .

The time period will be 2π√(L+a/2)/g But this is incorrect .

I would be grateful if somebody could help me with the problem.

 

[TSny]

I think the 1st picture is correct . In that case the plate is not rotating about the CM and the plate can be simply replaced by its CM .

Good.

Does the CM move along the arc of a circle? If so, where is the center of that circle?

 

[Simon Bridge]

Not sure which one correctly describes the situation

... well you cannot get the rest unless you get this right! How would you figure that out for yourself?

Ask yourself: do the strings change length during the motion?
Which diagram agrees with the answer above?

[edit] looks like I was beaten to the punch :)

 

[Tanya Sharma]

Good.

Does the CM move along the arc of a circle? If so, where is the center of that circle?

Hello TSny !

Yes...the CM moves along the arc of a circle . The center of the circle is the mid point of the length between the top end points of the string.

The blue dot represents the center .

Isn't the CM at a distance L+a/2 from the center?

 

[ehild]

Yes, the plate will not rotate as the lengths of the strings can not change can you prove it?
How did you get your result? What forces act on the plate when the strings make an angle with the vertical?

ehild

 

[TSny]

Hello TSny !

Yes...the CM moves along the arc of a circle . The center of the circle is the mid point of the length between the top end points of the string.

The blue dot represents the center .

Isn't the CM at a distance L+a/2 from the center?

No. The blue dot is not the center of the circle that the CM moves along. Keep thinking about this.

 

[Tanya Sharma]

No. The blue dot is not the center of the circle that the CM moves along. Keep thinking about this.

Please forgive me if my answer doesn't make sense . I am just guessing .

Is it the top end of the left string ?

Does the center changes alternately between the two top end points of the strings ?

 

[Tanya Sharma]

Yes, the plate will not rotate as the lengths of the strings can not change can you prove it?
How did you get your result? What forces act on the plate when the strings make an angle with the vertical?

ehild

Since the plate does not rotate about the CM ,then the mass of the plate can be considered to be at the CM .

I think of the system of string and plate as similar to a simple pendulum with a time period 2π√(L/g) .

I thought the CM was at a distance L+a/2 from the ceiling,so the time period was 2π√(L+(a/2)/g)

I feel I am making some fundamental mistakes.

The forces acting on the plate are the tensions from the strings and the weight .

 

[ehild]

"I think" does not mean anything in Physics. You have to prove your guess.

Prove that the plate can not rotate.

If so, show the equation for motion of the CM. What forces act on the plate if the strings make an angle with the vertical?

You can also use the analogy with a mathematical pendulum, but you need to show that the CM moves on a circle, and find the centre and radius of that circle.

ehild

 

[ehild]

Please forgive me if my answer doesn't make sense . I am just guessing .

Is it the top end of the left string ?

Does the center changes alternately between the two top end points of the strings ?

If the centre of a "circle" changes, it is not a circle.

What are the coordinates of the CM with respect to the blue dot when the strings make an angle θ with the vertical?


ehild

 

[Tanya Sharma]

Prove that the plate can not rotate.

The tensions in the two strings are providing equal and opposite torques about the CM and the line of action of weight passes through the CM .Since net torque is zero,plate does not rotate.

If so, show the equation for motion of the CM. What forces act on the plate if the strings make an angle with the vertical?

Let the tension in the left string be T1 and right string be T2 .Then T1=T2=T

In the radial direction , 2T = mgcosθ , where θ is the angle which the strings make with the vertical.

In the tangential direction net force = -mgsinθ

What are the coordinates of the CM with respect to the blue dot when the strings make an angle θ with the vertical?
ehild

The coordinates of CM with respect to the blue dot (i.e origin) are [(L+a/2)sinθ , -(L+a/2)cosθ]

Is it correct ?

 

[ehild]

The tensions in the two strings are providing equal and opposite torques about the CM

Are you sure? Why? Is not possible that the tensions are different in the strings?

Let the tension in the left string be T1 and right string be T2 .Then T1=T2=T

In the radial direction , 2T = mgcosθ , where θ is the angle which the strings make with the vertical.

What do you mean on radial and tangential direction? Would not it be easier to use horizontal and vertical directions?

Does the CM not move in the vertical direction?

In the tangential direction net force = -mgsinθ
The coordinates of CM with respect to the blue dot (i.e origin) are [(L+a/2)sinθ , -(L+a/2)cosθ]

Is it correct ?

No.

 

[Tanya Sharma]

If the tensions are different wouldn't the plate rotate about the CM ?

Are the coordinates of CM [Lsinθ , -(Lcosθ+a/2)] ?

 

[ehild]

If the tensions are different wouldn't the plate rotate about the CM ?

It would, but you can not be sure that the tensions are equal. As you can not be sure that the plate does not rotate. What is that you know for sure?

Are the coordinates of CM [Lsinθ , -(Lcosθ+a/2)]

Much better It is true if the y-axis points upward. So is it a circle, the CM moves on?

ehild

 

[Tanya Sharma]

It would, but you can not be sure that the tensions are equal. As you can not be sure that the plate does not rotate. What is that you know for sure?

I am sure only about one thing ; that I can not solve this problem :tongue:

Just kidding... that the strings are inextensible .

Much better It is true if the y-axis points upward. So is it a circle, the CM moves on?

ehild

No,the CM doesn't move in a circle about the blue dot .

But the CM does move in a circle . Right ?

 

[ehild]

I am sure only about one thing ; that I can not solve this problem :tongue:

Just kidding... that the strings are inextensible .

You can solve the problem.

Is it possible that the strings make different angles with the vertical if they are of the same length?

No,the CM doesn't move in a circle about the blue dot .

But the CM does move in a circle . Right ?

It is right. What are the centre and radius of the circle?

ehild

 

[Tanya Sharma]

What are the centre and radius of the circle?
ehild

Is (0,-a/2 ) center of the circle (purple) i.e point at a vertical distance a/2 downwards to the blue dot ?

In that case the radius is L .

Does that make sense ?

 

[ehild]

Is (0,-a/2 ) center of the circle (purple) i.e point at a vertical distance a/2 downwards to the blue dot ?

In that case the radius is L .

Does that make sense ?

It does...

ehild

 

[Tanya Sharma]

Thanks ehild...

What do you mean on radial and tangential direction? Would not it be easier to use horizontal and vertical directions?

Does the CM not move in the vertical direction?

The restoring force would be along the tangent to the arc i.e -mgsinθ.How would we come up with the restoring force by working with accelerations in horizontal and vertical directions ?

Again I am still not clear how did you and TSny straightaway knew that the CM moves in a circle .

I came up with the answer by a lot of incorrect guesses and manipulating the coordinates of the CM w.r.t the blue dot .

I incorrectly guessed that the blue dot was the center .From that coordinates of CM were found .Then I shifted the origin and tried to manipulate the coordinates so as to fit in the motion of CM along the arc .

But what if I didn't know that the CM moves along the arc .

What is the correct way to think about this problem ?

 

[ehild]

I did not know that the CM would move along a circle. But it came out ...

You can guess but before working with it, prove it is true.

I wrote up the relation between acceleration and forces for the horizontal and vertical components. Cancelled the tensions. Then I used the relation between angle and coordinates. Not a nice method, but I can not make too big mistakes.

ehild

 

[Tanya Sharma]

Could you explain how the tensions in the two strings will be unequal,yet no torque about the CM.

 

[ehild]

Could you explain how the tensions in the two strings will be unequal,yet no torque about the CM.

It can happen if the strings are not parallel.

ehild

 

[TSny]

Could you explain how the tensions in the two strings will be unequal,yet no torque about the CM.

When the strings are not vertical, can you see that the moment arms (lever arms) about the CM for the two tensions are not equal?

Can you determine at what angle θ one of the strings must go slack if the oscillations are large enough?

 

[TSny]

Thanks ehild...
Again I am still not clear how did you and TSny straightaway knew that the CM moves in a circle .

Consider the quadrilateral whose corners are at the ends of the two strings. So, two sides are the strings, one side is the line segment connecting the upper ends of the string, and the 4th side is the line segment connecting the lower ends of the string. What kind of quadrilateral is this (assuming both strings are taut)? If the top of the quadrilateral remains horizontal, what can you say about the bottom of the quadrilateral?

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]

When the strings are not vertical, can you see that the moment arms (lever arms) about the CM for the two tensions are not equal?

But the strings remain parallel throughout the motion. Right ?

The green and blue lines are the moment arm .

How should I convince myself that they are unequal ?

 

[Tanya Sharma]

Consider the quadrilateral whose corners are at the ends of the two strings. So, two sides are the strings, one side is the line segment connecting the upper ends of the string, and the 4th side is the line segment connecting the lower ends of the string. What kind of quadrilateral is this (assuming both strings are taut)? If the top of the quadrilateral remains horizontal, what can you say about the bottom of the quadrilateral?

The quadrilateral is a parallelogram . The bottom side will remain parallel to the top side.

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.

Did you pick anyone point and analysed its trajectory and then deduced that the CM should also have that trajectory ?

If yes,even in that case ,how did you come up that the trajectory is an arc ?

 

[TSny]

Can you see that your blue line is longer than your green line?

It helps to draw the plate with a larger vertical dimension.

 

[TSny]

The quadrilateral is a parallelogram . The bottom side will remain parallel to the top side.

Yes.

Did you pick anyone point and analysed its trajectory and then deduced that the CM should also have that trajectory ?

If yes,even in that case ,how did you come up that the trajectory is an arc ?

The trajectory of the point of attachment of one of strings to the plate should be easy to see.

 

[Tanya Sharma]

The trajectory of the point of attachment of one of strings to the plate should be easy to see.

Excellent thinking

Can you see that your blue line is longer than your green line?

Is it right to say that as the plate moves towards right upwards ,the moment arm of tension in the left string decreases (tension T1 increases) and the moment arm of the tension in the right string increases (tension T2 decreases ) such that T₁r₁ = T₂r₂ ?

 

[TSny]

Is it right to say that as the plate moves towards right upwards ,the moment arm of tension in the left string decreases (tension T1 increases) and the moment arm of the tension in the right string increases (tension T2 decreases ) such that T₁r₁ = T₂r₂ ?

That's right. You can see the angle θ at which the moment arm on the left goes to zero. So, what happens to T₂ at that angle?

 

[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.