M.50 Five particles, each of mass m, lie on a a frictionless horizontal plane. T

Question

M.50 Five particles, each of mass m, lie on a a frictionless horizontal plane. The position velocity, and acceleration of each particle at the instant t = 0 are presented in the table below. BLOCK AN Position VELOCITY ACCELERATION (2,4) (0,3) ( 2) (4,1) (-1, 1) ( 5) (6,4) (-1, 2) (0,0) (8, 1) (-1, 1) ( 2) (10,4) (-1, -1) 5 (a) Sketch the positions of the particles on the plane and indicate with an arrow the direction in which each is moving at t = 0. (b) (i) Determine the location of the centre of mass of the system at the instant t = 0, and (ii) mark its position on the sketch produced in (a). (c) (i) Determine the velocity of the centre of mass of the system at the instant t = U, and (ii) indicate its direction with an arrow at the point obtained in (b). (d) (i) Determine the acceleration of the centre of mass of the system at the instan t = 0). (ii) Comment on the net external force acting on this five-particle system.

Explanation / Answer

1) You can draw this on your own.

2) location of center of mass

xcm = m1x1 + m2x2 + m3x3 + m4x4 + m5x5 / m1 + m2 + m3 + m4 + m5

as all particles have same mass,

xcm = 2m + 4m + 6m + 8m + 10m / m+m+m+m+m

xcm = 30m / 5m

xcm = 6

Similarly, ycm = 4m + 1m + 4m + 1m + 4m / m+m+m+m+m

ycm = 14m / 5m

ycm = 2.8

(location)cm = (6, 2.8)

c) velocity of center of mass = 0m - 1m - 1m - 1m -1m / m+m+m+m+m ( x- direction)

vcm-x = -4m / 5m = - 0.8

vcm-y = 1.2

vcm = sqrt( 0.82 + 1.22 ) (Total magnitude)

vcm = 1.4422 m/s

d) Similarly, do for acceleration,

acm-x = 1 m / 5m = 0.2

acm-y = 0.2

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.