System of Particles and Rotational Motion
- A force →F = αî + 3ĵ + 6̂k is acting at a point →r = 2î - 6ĵ - 12̂k . The value of α for which angular momentum about origin is conserved is :
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From Newton's second law for rotational motion,
(2î - 6ĵ - 12&770;k) × (αî + 3ĵ + 6̂k) = 0
Solving we get α = –1Correct Option: D
From Newton's second law for rotational motion,
(2î - 6ĵ - 12&770;k) × (αî + 3ĵ + 6̂k) = 0
Solving we get α = –1
- Point masses m1 and m2 are placed at the opposite ends of a rigid rod of length L, and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point P on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity ω0 is minimum, is given by :
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Work required to set the rod rotating with angular velocity ω0
K.E. = (1/2)Iω²
Work is minimum when I is minimum.
I is minimum about the centre of mass
So, (m1) (x) = (m2) (L – x)
or, m1x = m2L – m2x∴ x = m2L m1 – m2 Correct Option: C
Work required to set the rod rotating with angular velocity ω0
K.E. = (1/2)Iω²
Work is minimum when I is minimum.
I is minimum about the centre of mass
So, (m1) (x) = (m2) (L – x)
or, m1x = m2L – m2x∴ x = m2L m1 – m2
- A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N ?
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Given, mass of cylinder m = 3kg
R = 40 cm = 0.4 m
F = 30 N ; α = ?
As we know, torque τ = Iα
F × R = MR²αα = F × R MR² α = 30 × (0.4) or, α = 25red/² 3 × (0.4)² Correct Option: B
Given, mass of cylinder m = 3kg
R = 40 cm = 0.4 m
F = 30 N ; α = ?
As we know, torque τ = Iα
F × R = MR²αα = F × R MR² α = 30 × (0.4) or, α = 25red/² 3 × (0.4)²
- The angular speed of an engine wheel making 90 revolutions per minute is
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Number of revolutions made by the engine wheel (n) = 90 per minute.
Angular speed of the engine wheel(ω) = 2πn = 2π × 90 = 3 π rad/s. 60 60 Correct Option: B
Number of revolutions made by the engine wheel (n) = 90 per minute.
Angular speed of the engine wheel(ω) = 2πn = 2π × 90 = 3 π rad/s. 60 60
- Three masses are placed on the x-axis : 300 g at origin, 500 g at x = 40 cm and 400 g at x = 70 cm. The distance of the centre of mass from the origin is :
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Xcm = m1x1 + m2x2 + m3x3 m1 + m2 + m3 Xcm = 300 × (0) + 500 × (40) + 400 × (70) 300 + 500 + 400 Xcm = 500 × 40 + 400 × 70 1200 Xcm = 50 + 70 = 120 = 40 cm 3 3 Correct Option: A
Xcm = m1x1 + m2x2 + m3x3 m1 + m2 + m3 Xcm = 300 × (0) + 500 × (40) + 400 × (70) 300 + 500 + 400 Xcm = 500 × 40 + 400 × 70 1200 Xcm = 50 + 70 = 120 = 40 cm 3 3
