Intuition for COM
Trending Questions
- a12, −b12
- −a12, −b12
- −a12 , b12
- a12, b12
- S
- R
- P
- Q
lf all the particles of a system lie in X-Y plane, is it necessary that the centre of mass be in X-Y plane?
Never
Some times
Not theoretically but practically
Yes
Four particles, each of mass m, are placed at the corners of a square of side a in the x-y plane. If the origin of the coordinate system is taken at the point of intersection of the diagonals of the square, the coordinates of the centre of mass of the system are
(-a, a)
(a, -a)
(0, 0)
(a, a)
- 8×10−2J
- 16×10−2J
- 4×10−4J
- 1.6×10−3J
Four particles A, B, C and D having masses m, 2m, 3m and 4m respectively are placed in order at the corners of a square of side a. Locate the centre of mass
,
None of these
,
Four particles of masses 1 kg, 2 kg, 3 kg and 4 kg are placed at the corners of a square of side 2 m in the x-y plane as shown in the figure. If the origin of the coordinate system is taken at the mass of 1 kg, the (x, y) co-ordinates of the centre of mass are
(1 m, 75 m)
(2 m, 75 m)
(3 m, 75 m)
(4 m, 75 m)
- 4×10−4 m
- 5×10−4 m
- 4.5×10−4 m
- 3×10−4 m
Look at the drawing given in the figure, Which has been drawn with ink of uniform line thickness. The mass of ink used to draw each of the two inner circles, and each of the two line segments is m. The mass of the ink used to draw the outer circle is 6m. The coordinates of the centres of the different parts are : outer circle (0, 0), left inner circle (-a, a), right inner circle (a, a), vertical line (0, 0) and horizontal line (0, - a). The y – coordinate of the centre of mass of the ink in this drawing is ?
a8
a10
a12
a3
- (2, 2)
- (4, 2)
- (6, 6)
- (1, 2)
Four particles, each of mass m, are placed at the corners of a square of side a in the x-y plane. If the origin of the coordinate system is taken at the point of intersection of the diagonals of the square, the coordinates of the centre of mass of the system are
(a, a)
(a, -a)
(-a, a)
(0, 0)
- M & 2M
- 2F
- M & 3M
- 3F
- 2M & 2M
- 4F
- a2(3i+5j)
- a2(3i+j/√3)
- a2(3i+j)
- a2(i+1√3)