- #1
Saitama
- 4,243
- 93
Homework Statement
Homework Equations
The Attempt at a Solution
I don't have the slightest idea on how should i go on making equations, any help would be appreciated.
Infinitum said:Try using the fact that the horizontal distance between the points OA1, A1A2, A2A3 remain in the same ratio, and hence their velocities.
Pranav-Arora said:How would i go on finding OA1, A1A2 and A2A3?
Pranav-Arora said:How would i go on finding OA1, A1A2 and A2A3?
Infinitum said:You can't find their exact values, but you sure can find their ratios. Let the first rhombus have length 5l, second 2l, and third 2l. Now, when the outermost hinge is pulled, all their angles will change by the same amount. So you have the distance ratio as...?
Hint : A tiny bit of trigonometry will help
I like Serena said:You're good in drawings aren't you?
What if you make a drawing with the construction fully folded, meaning A1=A2=A3.
And another one with the construction fully extended.
Suppose we measure starting from the wall.
What are the distances to the wall of each point in both cases?
Pranav-Arora said:For the second case, the distances will be 10, 6 and 4. I am not able to understand the first case.
I like Serena said:Those are the distances between the points.
But you need the distances to the wall...?
Pranav-Arora said:Seems like i am doing something wrong, i am getting the distance ratio as 5:3:2. :uhh:
Infinitum said:Uh yes, I meant the distance ratios from the origin will remain constant.
Pranav-Arora said:Oops, sorry, they are 10, 16 and 20.
Pranav-Arora said::tongue:
So now what about the velocity? I think i have reached the answer, its d) but i still don't get it why the distance ratio remains constant, how did you come to know that we need to solve it using distance ratio?
Pranav-Arora said::tongue:
So now what about the velocity? I think i have reached the answer, its d) but i still don't get it why the distance ratio remains constant, how did you come to know that we need to solve it using distance ratio?
I like Serena said:Much better! :)
So suppose we fold the construction completely.
The distances to the wall will then be 0, 0, and 0.
Let's call this t=0.
Now we extend it such that A3 has constant velocity V.
The means that at some time t the distance of A3 is Vt=20.
What was the distance to the wall of A2 at that time?
I like Serena said:Good!
A hinged rhombus is a four-sided polygon with opposite sides of equal length and four angles of equal measure. It is hinged at one of its vertices, allowing it to rotate and change shape.
To find the velocity of a hinged rhombus, you will need to know the length of the sides and the angle of rotation. Then, you can use the formula v = Δθ/Δt to calculate the angular velocity, which can be converted to linear velocity by multiplying by the radius of the hinged vertex.
Linear velocity is the rate of change of an object's position in a straight line, while angular velocity is the rate of change of an object's angular position around a fixed point. In the case of a hinged rhombus, linear velocity refers to the speed at which the hinged vertex is moving in a straight line, while angular velocity refers to the speed at which the rhombus is rotating.
Yes, the velocity of a hinged rhombus can be negative. This would occur if the rhombus is rotating in a clockwise direction, resulting in a negative angular velocity. However, the linear velocity at the hinged vertex may still be positive if it is moving in a straight line away from the hinged point.
One application could be in engineering, where hinged rhombuses are commonly used in mechanisms such as folding tables and doors. By calculating the velocity, engineers can ensure that these mechanisms are moving at the desired speed and are functioning properly. Other applications may include robotics, where hinged rhombuses are used in robotic arms and joints, and in physics experiments to study rotational motion.