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The three-dimensional motion of a particle is defined by the position
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Vector Mechanics for Engineers: Dynamics
- A revolving slotted arm OA moves a bearing P in a fixed curve shape: r(θ) = Cθ. r = radial distance to O θ = angle the arm OA makes in the x-direction C = known constant Arm OA begins from rest when θ = π/4 and rotates counterclockwise with constant angular acceleration d2θ/dt2 = α. Part 1: Calculate v(t) (the velocity vector) of the P bearing as a function of t. Show result with respect to C, α, and the unit vectors ur (r-direction) and uθ (θ-direction). Part 2: Calculate a(t) (acceleration vector) of the P bearing as a function of t. Show result with respect to C, α, and the unit vectors ur (r-direction) and uθ (θ-direction). Part 3: Calculate magnitude v of the velocity and the magnitude a of the acceleration of P when the angle of the slotted arm is θ = 3π/4.arrow_forward5 (a)The x-coordinate of a particle in curvilinear motion is given by x = 3.7t3 - 4.5t where x is in feet and t is in seconds. The y-component of acceleration in feet per second squared is given by ay = 1.7t. If the particle has y-components y = 0 and vy = 3.4 ft/sec when t = 0, find the magnitudes of the velocity v and acceleration a when t = 5.9 sec. Sketch the path for the first 5.9 seconds of motion, and show the velocity and acceleration vectors for t = 5.9 sec.arrow_forwardAt time t = 17 s, the velocity of a particle moving in the x-y plane is v = 0.14i +2.92j m/s. By time t = 17.13 s, its velocity has become 0.01i + 2.82j m/s. Determine the magnitude day of its average acceleration during this interval and the angle made by the average acceleration with the positive x-axis. The angle is measured counterclockwise from the positive x-axis. Answers: dav 0= i 0.546 i 37.56 m/s²arrow_forward
- 6(a) The y-coordinate of a particle in curvilinear motion is given by y = 4.5t3 - 5.9t, where y is in inches and t is in seconds. Also, the particle has an acceleration in the x-direction given by ax = 9.6t in./sec2. If the velocity of the particle in the x-direction is 12.9 in./sec when t = 0, calculate the magnitudes of the velocity v and acceleration a of the particle when t = 1.4 sec. Construct v and a in your solution.arrow_forwardAt time t = 12 s, the velocity of a particle moving in the x-y plane is v=0.08i + 2.28j m/s. By time t = 12.12 s, its velocity has become -0.08i + 2.14j m/s. Determine the magnitude day of its average acceleration during this interval and the angle 0 made by the average acceleration with the positive x-axis. The angle is measured counterclockwise from the positive x-axis. Answers: dav 0= Mi i m/s² Oarrow_forwardA A particle is moving along a curve. Its x position is described by x = t² m while its y position is given by the function y = 2 + t4 m. Determine the following: a. Sketch an x-y plot of the particle's path, showing the starting position (at t = 0) of and the direction taken by the particle b. At t= 1 sec, determine the velocity and acceleration vectors of the particle. (Express in Cartesian vector form) c. At t = 1 sec, determine the normal and tangential components of the acceleration. Show the vector diagram of the components. d. Determine the magnitude of the acceleration at t = 1 sec.arrow_forward
- answer a b c onlyarrow_forwardA particle which moves with curvilinear motion has coordinates in millimeters which vary with the time t in seconds according to x = 4.8t2 - 4.2t and y = 3.7t2 - t/2.7. Determine the magnitudes of the velocity v and acceleration a and the angles which these vectors make with the x-axis when t = 2.9 s. Answers: When t = 2.9 s, V = i mm/s, i i mm/s?, 0= i a =arrow_forwardYour answer is partially correct. A particle which moves with curvilinear motion has coordinates in millimeters which vary with the time t in seconds according to x = 1.5t2 - 5.2t and y = 2.3t2 - t/5.8. Determine the magnitudes of the velocity v and acceleration a and the angles which these vectors make with the x-axis when t= 1.4 s. Answers: When t = 1.4 s, mm/s, O, = i V = 5.51757 46.4108 4.3512 mm/s?, e, a = 46.4108arrow_forward
- A new discovered asteroid were observed by astrophysicist. It is said to be moving along a circular path of radius 500 km. If it is revolving with an angular velocity of (4t3) rad/s, where t is in seconds, find the magnitude of the asteroid's velocity and acceleration when 0 = 45°. Note that Cr = 0° when t=0.arrow_forwardA ball (shown in red color) is constrained to move in a circular path and it is also force to move downwards by the slider (shown in blue color) at a constant velocity of 0.5 m/s. Find the magnitude of acceleration of the ball at the instant when 0 = 60°. Show all necessary diagrams. Slider 2 m 0.5 m/sarrow_forwardSolve it correctlyarrow_forward
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