4.Arod of length Lo = 1.50 m moves along the horizontal direction: The rod makes an angle 80 25.09 with the x axis of its own moving reference frame S' . (a) H...

A uniform rod of mass $1.0 \mathrm{kg}$ and length $2.0 \mathrm{m}$ is free to rotate about one end (see the following figure). If the rod is released from rest at an angle of $60^{\circ}$ with respect to the horizontal, what is the speed of the tip of the rod as it passes the horizontal position?

Birds start with the diagram we have. This is X success and this is why excess and this is our rot. The angle here is 30 degrees. This is of length L And this is still length l X from here here. And this is the length away from origin to this point. And this is the direction of motion. So the X component of length there is a contracted of the rod in moving frame is hello X. That is an L cost 30 degrees. So we have a l s 2 m because 30 degrees. So, Alexis, 1.73 me to now the white continent of the length of the road in moving frame, that is l y is l sign 30 degree. So that will be 1 m now on the X component of proper length. Mhm of the rod? Yes. Next is game off. L X here. Gamma is the Lawrence factor. So we have Gamma is one upon route over one minus. We square upon C square. So we have one upon route over one minus B s 0.995 c square upon. We have C square. That will be 1 10.1 Now Ah, lettuce. Find the proper length of the road. That is L. P. And the X component. We have got my 0.1 and l x is 1.73 meters. There will be so 29.31 meters. No, Um when the road is moving with the relative speed, only the length parallel to its motion is affected, whereas the length perpendicular to the motion is constant. So the, um why continent mhm of proper length of the rod is crucial to this. So we have l. Y s 1 m. Therefore the proper length of the roadies route over and the X Square. This is the right component LP X square plus LP. Why square? So we have l P 17.31 m square plus 1 m square. This is all on the route, so the proper length of the road is 17.34 m. Now, the second one let as far be the angle made by the road with positive X access in. Come in, found third clockwise direction in proper fee frame. Yeah, so Alpha is then universe LP. Oh, why are Born L B X. So 10 inverse L. P. Y is 1 m L. P X is 17.31 m, so as far as 3.3 degrees. Therefore, the angle made by the roar with post positive X axis in counterclockwise direction in proper frame is a 3.3 degree.

In this question for the first part, we have to determine the proper length of T. Yeah. No. So the length of the road for particular to emotion is saved in a proper frame of fraud and the frame investor roadies movie. So the expression for perpendicular continent of the rod. Yes, mm. And why is according to L sign Data that there's a girl two Phil B. Y. Here this is the Y. Component of the problem and makes the X. Component of the length in our frame of reference is of course This is equation one. Um and the expression were Y. Component of the length in our frame of reference is I'm peter, this is a question too. So now, uh the relation between This two equation is uh huh. He executed two x. Why route over one minus the square upon C. Square. This is equation three. Here, C. Is the speed of light. So the proper length will be square root of the length of the square of the as the Y. And L. C. X. So. And he is going to right over the way square. Hello? And he checks this is the question. Sure. So we have um in equation too we have um have a good so and is two m Market slide by sign 30°.. This will be one m. That is according to can speak why. And in here in this equation we have route over and B. Y. Is. So um in this equation we have access seven point 1.73 m upon October 19 0.995 C square one. He is C square recorded too. was 17.3 meter putting the value at the accident. Do I in here? So 17.3 sq. This one m square is equality 17.4 m. therefore the proper link or stay rod. A 17.4. Mhm. Now, that's where the second fight, where we have to determine the orientation of angle improper frame. So the expression for angle of orientation is to then in verse at the white upon as the that is equal to an invoice and white upon gamma. Help X. Disses equation. But it's now from uh equation strong equation one and two. Let us substitute the value of then picked up tent it up for and why upon LX in patient five. So we have um it is able to then in birth and peter upon gamma. This is the question. Think so. Um So to to dash is equal to then in verse Dante to afghan one not born fruit over when when we square up on C. Right the thing that I live camera here. So we have um Then in verse then 30 degree. Multiplied over one The is 995 c. Where okay upon say square, so that it will be 3.3°.. Therefore the orientation orientation of angle in proper frame is 3.3 Degree.

Prince. This is the problem based book. Yeah, the Nativity. Here it is. Given moving Brought absurd. Toby having the length 2 m on an in orientation 30 degree. And with respect. Oh, direction or position. Oh, you Uh huh. If spittle fraud is 1995 c fire proper lynch and proper orientation Don't has been Obama is to find us root off one minus. We square by C Square. So it is Toby. Dr ST. Okay, so l one x is going toe. Elvin cause off the travel that is tooling toe Quite so 30 relief. That is 1.73 m and one by will be urban side off the tavern that it to side off 30 That has been rated. Yeah, And to access the proper, let alone exercise, which is equal toe and one x into gamma. That is 1.73 in tow. 10. So 17.3 m and two white is good toe and one x Alban by story. Yeah, that is 1 m. Mhm heads and you will be square. Root off had to work. Cisco that is 17.3 square plus bonus square. So it is Toby. 17.4 m theater, too, will be 10 and worse off and to buy upon two X that is Dannon worse off born upon 17.3. So it is equal to 3.30 to there, so thanks for watching it.