Answer:
Perpendicular to the line joining th points
Explanation:
The relative velocity of anypany two on link always normal to line joining of two points.
Question 2 The direction of linear velocity of any point on a link with respect to...
QUESTION 4 A link AB is shown below. The direction of linear velocities of points A and B at a particular instant are shown in the figure as well. If the instantaneous angular velocity of the link is w then the magnitude of the linear velocities of points A and B are, = 25 w and v= 1.86 - 1.154 and v, = 2.31 1A = 2.31 w and wg = 1.154 A = 1.86 w and Yg = 2.5...
QUESTION 5 Velocity direction of any point on a rigid body undergoing pure rotation is tangent to its path of motion. True False QUESTION 6
Determine the angular velocities (ω3 and ω4) and the linear velocity (magnitude and direction) of point P. The filled circles are the joints of the fixed link 1. Use the dimension of the link provided to determine the appropriate scale and to determine the remaining link lengths, if need be. 4 - 29 in W, - 2 rad/s cow
kinematics In this system, rotation of link AB is translated into linear movement of piston P. If the angular velocity of link AB is (o-1rad/sec), find the linear velocity of piston P (point D in the picture). Remember: There are multiple ways to solve this problem, one is to use relative velocity vector equation as we worked out in class. Or another method is to use the center of instantaneous zero velocity for link BD For this part of the...
c. It has tangential (linear) acceleration that points in the same direction as the velocity, and a decreasing centripetal acceleration d. It has tangential (linear) acceleration that point opposite the velosity, and a constant centripetal acceleration QUESTION 3 A 53.5 kg parachutist jumps from an airplane and falls to Earth with a drag force proportional to the square of the speed, R-CV. Take C -0.220 kg/m (with the parachute closed) and C - 22.0 kg/m (with the chute open). What...
Solve for the angular velocity of link AB in rad/s. You may assume point A is moving in the -y direction and point B moves in the x- direction. у A X TB/A 0.2m VA = 2 m/s 45° B VB
ME 322 Final Take Home Question Name Part 2. (20 points) In the following mechanism, OA 0.250 m. OB 0.273 m, link 2 is the input link. At the moment, θ,-45° , the angular velocity ω2 4 rad/s (CVV), the angular acceleration α2- 0.5 ras/s"(CCW), From position and velocity analysis, AB = 0.200 m, θ4-118.30, a4- 1.43 rad/s (CW), sliding velocity v33 0.9579 m/s, direction from B point to A. (choose your own method) Write the relative motion acceleration equation...
Question 3 5 pts Solve for the angular velocity of link AB in rad/s. You may assume point A is moving in the-y direction and point B moves in the x-direction. y А х 0.2m TBA va = 2 m/s 45° SB B
2) Determine the direction and magnitude of the electric field at the point P shown in the figure. The two charges are separated by a distance of 2a. Point P is on the perpendicular bisector of the line joining the charges, a distance x from the midpoint between them. Let a = 1.0 m, x-8.0 m, and Q= 5.0 uc -0
Please I need help with this questions. Thanks Question 1 Given that ver»,-. The point D lios directly to the left of O. T he direction of velocity of point D is PERPENDICULAR to which line (NOT drawn) 0.00 points out of 1.00 P Flag question no slip VR,a Select one: a. Oc c. BC d. CD Your answer is incorrect Question 2 Given that va, -O. The point D lies directly to the left of O. The direction of...