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#center of mass

38 public questions tagged with this topic.

Moment of Inertia and Radius of Gyration

A uniform square lamina of side 4m and mass 8kg has one corner at the origin and lies along the x- and y-axes. What is t

For a uniform square, the center of mass is at the centroid. Vertices: (0,0),(4,0),(0,4),(4,4). CM: X = 0+42 = 2, Y = 0+42 = 2. Position: (2,2)m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to (2,2). This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Moment of Inertia and Radius of Gyration

A 7kg object moves with a velocity of 3i^−4j^m/s. What is the magnitude of the velocity of its center of mass?

For a single object, the center of mass velocity equals the object’s velocity. Magnitude = (3)2+(−4)2 = 9+16 = 25 = 5m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 5.0 m/s. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Vector Product and Rotational Kinematics

A uniform triangular lamina has vertices at (0,0), (5,0), and (0,6). What is the x-coordinate of its center of mass?

For a uniform triangular lamina, the center of mass is at the centroid, the average of the vertices. X = 0+5+03 = 53≈1.67m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.67 m. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Vector Product and Rotational Kinematics

A 4kg block moves with a velocity of −2i^+3j^m/s. What is the magnitude of the velocity of its center of mass?

For a single object, the center of mass velocity equals the object’s velocity. Magnitude = (−2)2+(3)2 = 4+9 = 13≈3.6m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.6 m/s. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Vector Product and Rotational Kinematics

A uniform rectangular plate of dimensions 3m×5m and mass 12kg has one corner at (1,1) along the x- and y-axes. What is t

For a uniform rectangle, the center of mass is at the centroid. Vertices: (1,1),(4,1),(1,6),(4,6). CM: X = 1+32 = 2.5, Y = 1+52 = 3.5. Position: (2.5,3.5)m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to (2.5,3.5). This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Vector Product and Rotational Kinematics

Two particles of masses 3kg and 7kg are at (1,4) and (9,2) respectively. What is the distance of their center of mass fr

X = (3×1)+(7×9)3+7 = 3+6310 = 6.6. Y = (3×4)+(7×2)3+7 = 12+1410 = 2.6. Distance = (6.6)2+(2.6)2 = 43.56+6.76 = 50.32≈7.09m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 7.09 m. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Rotational Motion Applications

Three particles of masses 1kg, 5kg, and 10kg are at (0,2), (3,0), and (4,6) respectively. What is the x-coordinate of th

Formula: X = m1x1+m2x2+m3x3m1+m2+m3. Masses: 1,5,10kg; x-coordinates: 0,3,4. X = (1×0)+(5×3)+(10×4)1+5+10 = 0+15+4016 = 5516≈3.44m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.44 m. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Rotational Motion Applications

Three particles of masses 3kg, 6kg, and 9kg are located at (0,1), (2,2), and (4,0) respectively. What is the x-coordinat

Formula: X = m1x1+m2x2+m3x3m1+m2+m3. Masses: 3,6,9kg; x-coordinates: 0,2,4. X = (3×0)+(6×2)+(9×4)3+6+9 = 0+12+3618 = 4818 = 2.67m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.67 m. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Rotational Motion Applications

A 6kg object moves with a velocity of 3i^−2j^m/s. What is the magnitude of the velocity of its center of mass?

For a single object, the center of mass velocity equals the object’s velocity. Magnitude = (3)2+(−2)2 = 9+4 = 13≈3.6m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.6 m/s. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Rotational Motion Applications

Two particles of masses 4kg and 6kg are at (2,0) and (0,3) respectively. What is the distance of their center of mass fr

X = (4×2)+(6×0)4+6 = 810 = 0.8. Y = (4×0)+(6×3)4+6 = 1810 = 1.8. Distance = (0.8)2+(1.8)2 = 0.64+3.24 = 3.88≈1.97m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.97 m. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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