Acceleration of a point on square

In summary, the homework statement states that the velocity of the four rods will be the same, and that the work that will be done on the square is four Macm.
  • #1
Tanya Sharma
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Homework Statement



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2. Homework Equations [/B]

The Attempt at a Solution



The origin is placed to the right of point P and left is taken to be the positive direction.Let us call mass of each rod as M and the angle which rods make with the horizontal as θ .

Coordinates of A = (##x_P+\frac{L}{2}cosθ,\frac{L}{2}sinθ## )

Coordinates of B = (##x_P+\frac{3L}{2}cosθ,\frac{L}{2}sinθ## )

Coordinates of C = (##x_P+\frac{3L}{2}cosθ,-\frac{L}{2}sinθ##)

Coordinates of D = (##x_P+\frac{L}{2}cosθ,-\frac{L}{2}sinθ## )Now writing some constraint equations ,

##x_q = x_p + 2Lcosθ##

##v_q = v_p -2L(sinθ\dot{θ})##

##a_q = a_p -2L(cosθ\dot{θ}^2+sinθ\ddot{θ})##

##a_{cm} = \frac{1}{2}(a_p+a_q)##

The force applied on the square F= 4Macm . This force produces displacement of P ,xP in a very short time interval 't' . ##x_P = \frac{1}{2}a_Pt^2##

Work done on the square =FxP .

Now we will calculate the velocities of the CM of the rods .

##v_{A,x} = v_{P,x} - \frac{L}{2}sinθ\dot{θ}##
##v_{A,y} = \frac{L}{2}cosθ\dot{θ}##
##v_{B,x} = v_{P,x} - \frac{3L}{2}sinθ\dot{θ}##
##v_{B,y} = \frac{L}{2}cosθ\dot{θ}##

Similarly we can write speeds of the two bottom rods .

Magnitude of vertical component of velocity of CM of all four rods would be equal . Magnitude of horizontal component of velocity of CM of two right side rods would be equal whereas that of two left side rods would be equal.Angular speeds of all four rods would be same .

Now applying Work Energy theorem ,

$$4M\frac{1}{2}(a_p+a_q)(x_P) = 2(\frac{1}{2}M){(v_{P,x}- \frac{L}{2}sinθ\dot{θ})}^2+2(\frac{1}{2}M){(v_{P,x}- \frac{3L}{2}sinθ\dot{θ})}^2 + 4(\frac{1}{2}M){(\frac{L}{2}cosθ\dot{θ})}^2 + 4\frac{1}{2}(\frac{ML^2}{12})\dot{θ}^2$$

In the above expression we can put ##x_P = \frac{1}{2}a_Pt^2## , ##\dot{θ} = \ddot{θ}t ## and ##v_{P,x} = a_Pt##.

$$4M\frac{1}{4}(a_p+a_q)(a_Pt^2) = 2(\frac{1}{2}M){(a_Pt- \frac{L}{2}sinθ(\ddot{θ}t))}^2+2(\frac{1}{2}M){(a_Pt- \frac{3L}{2}sinθ(\ddot{θ}t))}^2 + 4(\frac{1}{2}M){(\frac{L}{2}cosθ(\ddot{θ}t))}^2 + 4\frac{1}{2}(\frac{ML^2}{12})(\ddot{θ}t)^2$$

Is it correct till here ? If yes, how do I eliminate ##\ddot{θ}## ?

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

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  • #2
aq can be also expressed by ##\ddot \theta##. Solve for ##\ddot \theta##, and determine aq with it.
 
  • #3
Is there another expression between aq and ##\ddot \theta## besides ##a_q = a_p -2L(cosθ\dot{θ}^2+sinθ\ddot{θ})## which I mentioned in the OP?
 
Last edited:
  • #4
Tanya Sharma said:
Is there another expression between aq and ##\ddot \theta## besides ##a_q = a_p -2L(cosθ\dot{θ}^2+sinθ\ddot{θ})## which I mentioned in the OP?

It is enough.
The velocity terms vanish. You can substitute sinθ and cosθ with their valus at t=0.
 
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Likes Tanya Sharma
  • #5
Thank you very much :oldsmile: .
 
Last edited:

Related to Acceleration of a point on square

1. What is acceleration?

Acceleration is the rate of change of an object's velocity over time. It is a vector quantity, meaning it has both magnitude and direction.

2. How is acceleration calculated?

Acceleration can be calculated by dividing the change in velocity by the change in time. It can also be calculated by dividing the net force acting on an object by its mass.

3. What is the formula for calculating acceleration?

The formula for acceleration is a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

4. How does acceleration affect an object's motion?

Acceleration can either increase or decrease an object's velocity, depending on its direction. If the acceleration is in the same direction as the velocity, the object will speed up. If the acceleration is in the opposite direction, the object will slow down.

5. How is acceleration represented graphically?

Acceleration is represented by a linear or curved line on a velocity vs. time graph. The slope of the line represents the acceleration, with a steeper slope indicating a greater acceleration.

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