= 30°. Force F of magnitude= 45° and a velocity vector v of magnitude 3.8 m/s and angle 02At the instant of the figure, a 1.6 kg particle P has a position vector r of magnitude 1.7 m and angle 011.6 N and angle 03= 30° acts on P. All three vectors lie in the xy plane. (Express your answers in vector form.)03(a) What is the angular momentum of the particle about the origin?kg · m2/s(b) What is the torque acting on the particle about the origin?デ=N: mAdditional Materials

Answered: Image /qna-images/answer/029673b3-fccc-47fa-a68c-6021a1bfa06d.jpg

Answered: At the instant of the figure, a 1.6 kg…

Similar questions

At a certain instant, the particle of mass m has the position and velocity shown in the figure, and it is acted upon by the force F. Determine its angular momentum about point O and the time rate of change of this angular momentum. Use m = 1.9 kg, a = 0.62 m, b = 0.99 m, c = 0.72 m, and F = 190 N and v= 3.6 m/s. V Answers: j+ i k) kg m²/s Ho= Ho = j+ k) N•m i m C i+ i+ i
Consider the force vector F=<1,1,1/2>. If the magnitude of the torque T=OP*F is equal to the area of the equilateral triangle formed by the origin, P, and (1,-1,1), then determine the acute angle formed by OP and F. Give your answer in degrees.
At latitude 30◦N the distance from the surface of the earth to the axis of rotation,r, is smaller than it is at the equator, by 13% (cos(30◦)≈0.87). If we assume that the angular momentum of a mass m(L=mrv) is conserved as the mass is moved from the equator to latitude 30◦N, how fast is its velocity,v, relative to the ground at 30◦N?
A particle of mass m in the figure below slides down a frictionless surface through height h and collides with a uniform vertical rod (of mass M and length d), sticking to it. The rod pivots about point O through an angle θ when it momentarily stops. Find θ in terms of m, M, g, h, d and various constants. (Please type answer no write by hend)
A projectile of mass .050 kg moves to the right with a speed of 4.0 m/s. The projectile strikes and sticks to the end of a rod of mass 0.8 kg, at length 1.2 m from the rod's centre. The rod may rotate freely around a stationary axle pinned through the rod's centre of mass. Find the magnitude of the angular velocity, ω, of the system right after the collision. Select one: a. 0.00 rad/s b. 0.27 rad/s c. 0.53 rad/s d. None of the other answers are correct e. 1.05 rad/s
A projectile of mass m moves to the right with a speed vi (see figure below). The projectile strikes and sticks to the end of a stationary rod of mass M, length d, pivoted about a frictionless axle perpendicular to the page through O. We wish to find the fractional change of kinetic energy in the system due to the collision. The moment of inertia of a rod is I = Md² and the moment of inertia of a particle is I = mr². 12 a. What is w, the angular speed of the system after the collision. mdvi Md²+1md² 12 b. What is the kinetic energy before the collision? Answer: KE¡ = m(v¡)². c. What is the kinetic energy after the collision? (0₁)² Answer: W= 1 m²d² Answer: KE₁ = = =+M₁2² +² md² 12 4 m O d M O 3
Figure 3 ft 4 ft 5 ft. < 1 of 1 If F₁ = {100i - 125j + 95k} lb and F₂ = {−210i + 280j + 100k} lb, determine the resultant moment produced by these forces about point O. Express the result as components of a Cartesian vector. (Figure 1) Enter the x, y, and components of the moment of force separated by commas. Mo = Submit V— ΑΣΦ 510, 1110,950 vec Provide Feedback Previous Answers Request Answer X Incorrect; Try Again; 3 attempts remaining lb. ft
à ở từ if a = 2.8i+ 5.7 j – 8.1k, = b = − 4.9î + 4.4ĵ + 3.8k, and What are (a) the x component, (b) the y component, and (c) the z component of 7 1.8î + 2.8ĵ + 4.6k . (d) Calculate the angle between and the positive z axis. (e) What is the component of a along the direction of b? (f) What is the magnitude of the component of a perpendicular to the direction of b but in the plane of a and b? =
A child's top is held in place upright on a frictionless surface. The axle has a radius of r 3.21 mm. Two strings are wrapped around the axle, and the top is set spinning by T T applying T = 2.40 N of constant tension to each string. If it takes 0.590 s for the string to unwind, how much angular 2r momentum L does the top acquire? Assume that the strings do not slip as the tension is applied. R SP 9.15 x10¬3 kg-m² L = S Point P is located on the outer surface of the top, a distance h = 35.0 mm above the ground. The angle that the outer surface of the top makes with the rotation axis of the top is 0 = 24.0°. If the final tangential speed v, of point P is 1.45 m/s, what is the top's moment of inertia I? T 2r T 2.42 x10-4 kg-m? = Incorrect
A projectile of mass .050 kg moves to the right with a speed of 4.0 m/s. The projectile strikes and sticks to the end of a rod of mass 0.8 kg, at length 1.2 m from the rod's centre. The rod may rotate freely around a stationary axle pinned through the rod's centre of mass. Find the magnitude of the angular velocity, ω, of the system right after the collision.
A merry-go-round is a playground ride that consists of a large disk mounted to that it can freely rotate in a horizontal plane. The merry-go-round shown is initially at rest, has a radius R = 1.3 meters, and a mass M = 251 kg. A small boy of mass m = 42 kg runs tangentially to the merry-go-round at a speed of v = 1.2 m/s, and jumps on. Randomized Variables R = 1.3 metersM = 251 kgm = 42 kgv = 1.2 m/s a)Calculate the moment of inertia of the merry-go-round, in kg ⋅ m2. b) Immediately before the boy jumps on the merry go round, calculate his angular speed (in radians/second) about the central axis of the merry-go-round. c) Immediately after the boy jumps on the merry go round, calculate the angular speed in radians/second of the merry-go-round and boy.
A merry-go-round is a playground ride that consists of a large disk mounted to that it can freely rotate in a horizontal plane. The merry-go-round shown is initially at rest, has a radius R = 1.5 meters, and a mass M = 251 kg. A small boy of mass m = 41 kg runs tangentially to the merry-go-round at a speed of v = 1.8 m/s, and jumps on.Randomized VariablesR = 1.5 metersM = 251 kgm = 41 kgv = 1.8 m/s (a) Calculate the moment of inertia of the merry-go-round, in kg ⋅ m2. Part (b) Immediately before the boy jumps on the merry go round, calculate his angular speed (in radians/second) about the central axis of the merry-go-round.(c) Immediately after the boy jumps on the merry go round, calculate the angular speed in radians/second of the merry-go-round and boy.(d) The boy then crawls towards the center of the merry-go-round along a radius. What is the angular speed in radians/second of the merry-go-round when the boy is half way between the edge and the center of the merry go round?(e) The…

SEE MORE QUESTIONS