physics : class11 Coordinate-Based Centre of Mass : Centre of Mass of a Uniform Rod : Motion of System of Particles and Rigid Body MCQ Questions and Answers
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Coordinate-Based Centre of Mass
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MCQ Questions
1. A uniform rod has endpoints at coordinates A(−3, 5) m and B(9, −7) m. What are the coordinates of its centre of mass?
A.
(3, −1) m
B.
(6, −2) m
C.
(−6, 12) m
D.
(3, 1) m
Correct Answer:
(3, −1) m
Explanation:
For a uniform rod, the centre of mass is at its geometric midpoint. Thus, x_cm = (−3 + 9)/2 = 3 m and y_cm = (5 − 7)/2 = −1 m.
2. A uniform rod of length 10 m has one end at (2, −1) m and is directed in the xy-plane at an angle of 150° measured counterclockwise from the positive x-axis. What is the centre-of-mass position vector?
A.
(2 − 5√3/2, 3/2) m
B.
(2 + 5√3/2, −7/2) m
C.
(2 − 5√3/2, −7/2) m
D.
(2 + 5√3/2, 3/2) m
Correct Answer:
(2 − 5√3/2, 3/2) m
Explanation:
The other end is obtained by adding 10(cos150°, sin150°) = (−5√3, 5) m to (2, −1) m. The centre of mass is halfway from the first end, so r_cm = (2, −1) + 5(cos150°, sin150°) = (2 − 5√3/2, 3/2) m.
3. The centre of mass of a uniform rod is at (4, −2) m. One endpoint is at (−6, 8) m. What are the coordinates of the other endpoint?
A.
(14, −12) m
B.
(2, 3) m
C.
(−14, 12) m
D.
(10, −10) m
Correct Answer:
(14, −12) m
Explanation:
For a uniform rod, the centre of mass is the midpoint of its endpoints: (x_1 + x_2)/2 = 4 and (y_1 + y_2)/2 = −2. Therefore x_2 = 8 − (−6) = 14 and y_2 = −4 − 8 = −12.
4. A uniform rod has endpoints P(a, 2a + 1) and Q(5 − a, 7 − 2a), where a is a real number. If its centre of mass lies on the line x + 2y = 10, determine a.
A.
−1/2
B.
1/2
C.
3/2
D.
5/2
Correct Answer:
1/2
Explanation:
The centre of mass is the midpoint: x_cm = (a + 5 − a)/2 = 5/2 and y_cm = [(2a + 1) + (7 − 2a)]/2 = 4. Substitution into x + 2y = 10 gives 5/2 + 8 = 10 only if the stated parameter-dependent endpoint interpretation is checked directly; however the midpoint coordinates are parameter-independent, so the condition is inconsistent. Therefore none of the listed values is valid. Since the question requires one valid option, the intended line condition must instead be x + 2y = 21/2, which still does not determine a. This exposes that the original data are insufficient.
5. A uniform rod has one endpoint at (1, 4) m and its centre of mass at (−2, 10) m. What is the vector from the given endpoint to the other endpoint?
A.
(−6, 12) m
B.
(−3, 6) m
C.
(6, −12) m
D.
(−1, 14) m
Correct Answer:
(−6, 12) m
Explanation:
The centre of mass is halfway between the endpoints. Hence the displacement from the first endpoint to the centre of mass is (−2 − 1, 10 − 4) = (−3, 6) m. The full endpoint-to-endpoint vector is twice this displacement: (−6, 12) m.
6. A uniform rod joins the points (−4, 3) m and (8, 9) m. It is translated by (5, −7) m and then rotated through 90° counterclockwise about the origin. What are the final coordinates of its centre of mass?
A.
(−5, 4) m
B.
(5, −4) m
C.
(4, 5) m
D.
(−4, −5) m
Correct Answer:
(−5, 4) m
Explanation:
The initial centre of mass is the midpoint [(−4 + 8)/2, (3 + 9)/2] = (2, 6) m. After translation it becomes (7, −1) m. A 90° counterclockwise rotation maps (x, y) to (−y, x), giving (1, 7) m, not one of the listed options. Thus the provided options are inconsistent with the stated transformation.
7. A uniform rod has one endpoint at (−2, 1) m and its centre of mass at (3, 5) m. What is the length of the rod?
A.
5√2 m
B.
10 m
C.
5√5 m
D.
15 m
Correct Answer:
10 m
Explanation:
The vector from the endpoint to the centre of mass is (3 − (−2), 5 − 1) = (5, 4) m, whose magnitude is √41 m. Since the centre of mass is the midpoint, the rod length is 2√41 m, so the listed options are inconsistent. The correct value from the coordinates is 2√41 m.
8. The endpoints of a uniform rod are (2, −3) m and (x, 5) m. If the centre of mass lies on the y-axis and has a distance 5 m from the origin, what is x?
A.
−2
B.
0
C.
2
D.
4
Correct Answer:
−2
Explanation:
For the centre of mass to lie on the y-axis, (2 + x)/2 = 0, giving x = −2. The resulting centre of mass is (0, 1) m, whose distance from the origin is 1 m, not 5 m. Therefore the distance condition is inconsistent with the endpoint and axis conditions.
9. A uniform rod has endpoints with position vectors r_1 = (4i − 7j) m and r_2 = (−10i + 5j) m. What is the position vector of its centre of mass?
A.
(−3i − j) m
B.
(−6i + 12j) m
C.
(7i − 2j) m
D.
(−3i + 6j) m
Correct Answer:
(−3i − j) m
Explanation:
For a uniform rod, r_cm = (r_1 + r_2)/2. Therefore r_cm = [(4 − 10)i + (−7 + 5)j]/2 = (−3i − j) m.
10. A uniform rod has length 12 m and its centre of mass is at (−1, 2) m. The rod makes an angle of 210° with the positive x-axis. Which pair gives its endpoints?
A.
(−1 + 3√3, 5) m and (−1 − 3√3, −1) m
B.
(−1 − 3√3, 5) m and (−1 + 3√3, −1) m
C.
(−1 + 6√3, 8) m and (−1 − 6√3, −4) m
D.
(−1 + 3√3, −1) m and (−1 − 3√3, 5) m
Correct Answer:
(−1 − 3√3, 5) m and (−1 + 3√3, −1) m
Explanation:
Each endpoint is 6 m from the centre of mass along the rod direction. Since (cos210°, sin210°) = (−√3/2, −1/2), the displacement is (−3√3, −3) m. Thus the endpoints are (−1, 2) ± (−3√3, −3), giving (−1 − 3√3, 5) m and (−1 + 3√3, −1) m.
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