In the design of a timing mechanism, the motion of pin P in the fixed circular slot is controlle

step 1 In this question, we have this setup. So there's a guide A, there's a pin P, and then we are giv

In the design of a timing mechanism, the motion of pin P in...

Key Concepts

Tangential Acceleration

This acceleration component acts along the tangent to the path and is responsible for changes in the magnitude of the velocity. It is determined by the rate at which speed increases or decreases, and is calculated as the derivative of speed with respect to time. In any problem where the speed of a moving object along a curved path is not constant, understanding tangential acceleration is essential.

Normal (Centripetal) Acceleration

This is the component of acceleration directed radially inward toward the center of the circular path. It quantifies the rate of change of the direction of velocity and is mathematically defined as the square of the speed divided by the radius (a_n = v²/r). Understanding this concept is critical when analyzing motions on curved paths.

Circular Motion Kinematics

This concept involves analyzing the motion of an object moving along a circular path. It includes understanding how displacement, velocity, and acceleration behave when the motion is constrained to a circle. Key kinematic relationships, such as the connection between linear velocity and angular speed (v = ?r), form the basis of this analysis.

Decomposition of Acceleration

In curvilinear (including circular) motion, the total acceleration vector can be split into two perpendicular components: the tangential acceleration, which is parallel to the motion and responsible for changes in speed, and the normal (centripetal) acceleration, which points toward the center of the path and is responsible for changing the direction of the velocity vector. This decomposition is fundamental in analyzing non-uniform circular motion.

Transcript

00:01 In this question, we have this setup.

00:03 So there's a guide a, there's a pin p, and then we are given that the guide a moves starts from rest.

00:23 It accelerates uniformly to the midway, to the midpoint of the curve track.

00:40 And then the speed is the vmax is 175 m per second and then after that it decelerates as it goes up okay and we want to find the normal component of the acceleration and the tangential component of the acceleration of pin p after moving 30 degrees from the start and then from the setup we are also given that like this starting point this starting point is 20 degrees from the from the vertical okay so yeah this is not quite accurate but yeah so it starts from 20 degrees and then you move 30 degrees okay so the pin p will be somewhere here and so the total angle from the vertical would be 50 degrees.

01:54 Okay, but first we need to find the acceleration of guide a, okay, that vertical upward acceleration for guide a.

02:05 And because the acceleration of guy a needs to be consistent with the normal component and the tangential components of the acceleration of the acceleration of of pin p okay so solution right so you are given that at halfway point the v max is 175 m per second then the distance traveled by the pin or distance traveled by a is equal to row sine 70 degrees in the radius row is 250 mm and so you calculate this you get 235 mm and then if you use using b squared equals to v0 square plus 2a delta y here our word is positive okay so and initial velocity is 0 so we can find acceleration to be v squared over 2 delta y and 175 square divided 2 times 2355 calculate the acceleration to be 65 .2 m m the second square i'm going to call this a y then you need to find the speed of the pin at 50 degrees okay so after moving 30 degrees then the angle from the vertical would be 50 degrees okay so you can imagine like this is the situation okay so this is the motion of pin p and then and initially it is at here okay so we want to find this distance y prime then for guide a is going to move as some v prime okay but on the on the pin is going to have a on the curve track a pin p will have a tangential velocity that looks like this i'm going to call this b okay all right so here is 20 degrees 30 degrees and 40 degrees okay okay i want to find y prime so y prime is equal to row sine 70 degrees minus row side 40 degrees okay so so this so this is row okay and along the curve try is always row so you can see that there's a right angle triangle here that connects these three points right in the triangle okay yeah so y prime is going to be row sign 70 degrees minus row sign 40 degrees and after putting the numbers you get said before and 2 and then okay then can calculate v prime using to a y y prime square root okay is this case using the same formula as above here just that the initial velocity is zero okay so two times 65 .2 times 74 .2 then take the square roots we calculate this you get 98 .4 m.

06:35 Per second okay so from the diagram we have the tangential velocity is like this and then the v prime will be pointing like this so here here is actually 40 degrees okay so how do i know that because here is 40 okay and the tangential velocity is 90 degrees to the radius and here is 50 okay yeah so the diagram okay v prime is equal to v cosine 40 degrees okay so v goes to v prime by cosine 40 degrees okay putting in numbers and calculate the tangential velocity is 1 to 8 .4 m per second okay then you can here this you can calculate the normal component of the acceleration is just v square over row.

07:57 Okay, so 1 to 8 .4 square divided by 250.

08:02 You get 66 .0 mm per second square...

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