System of Particles and Rotational Motion
- 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
- 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
- Two discs are rotating about their axes, normal to the discs and passing through the centres of the discs. Disc D1 has 2 kg mass and 0.2 m radius and initial angular velocity of 50 rad s–1. Disc D2 has 4kg mass, 0.1 m radius and initial angular velocity of 200 rad s–1. The two discs are brought in contact face to face, with their axes of rotation coincident. The final angular velocity (in rad s–1) of the system is
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Given: m1 = 2 kg m2 = 4 kg
r1 = 0.2 m r2 = 0.1 m
w1 = 50 rad s–1 w = 200 rad–1
As, angular momentum, I1W1 = I2W2 = Constant∴ Wƒ = = 1 m1r²h1w1 + 1 m2r²h2w2 I1W1 + I2W2 2 2 I1 + I2 1 m1r²1 + 1 m2r²2 2 2
By putting the value of m1, m2, r1, r2 and solving we get = 100 rad s–1Correct Option: C
Given: m1 = 2 kg m2 = 4 kg
r1 = 0.2 m r2 = 0.1 m
w1 = 50 rad s–1 w = 200 rad–1
As, angular momentum, I1W1 = I2W2 = Constant∴ Wƒ = = 1 m1r²h1w1 + 1 m2r²h2w2 I1W1 + I2W2 2 2 I1 + I2 1 m1r²1 + 1 m2r²2 2 2
By putting the value of m1, m2, r1, r2 and solving we get = 100 rad s–1
- 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
