System of Particles and Rotational Motion
- When a mass is rotating in a plane about a fixed point, its angular momentum is directed along :
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By right hand screw, rule, the direction of Ĺ is ⊥ to the plane containing
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The mass is rotating in the plane, about a fixed point, thus this plane will contain and the direction of Ĺ , will be ⊥ to the this plane. Correct Option: A
By right hand screw, rule, the direction of Ĺ is ⊥ to the plane containing.
The mass is rotating in the plane, about a fixed point, thus this plane will contain and the direction of Ĺ , will be ⊥ to the this plane.
- A circular disk of moment of inertia It is rotating in a horizontal plane, its symmetry axis, with a constant angular speed ωi. Another disk of moment of inertia Ib is dropped coaxially onto the rotating disk. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed ωƒ. The energy lost by the initially rotating disk to friction is:
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By conservation of angular momentum,
Itωi =(It + Ib) ωƒ
where ωƒ is the final angular velocity of disks∴ ωƒ = 
It 
ωi It + Ib
Loss in ΔK.E., = Initial K.E. – Final K.E.= 1 Itωi² - 1 (It + Ib)ωƒ² 2 2 ⇒ Δ K = 1 Itωi² - 1 (It + Ib) It² ωi² 2 2 (It + Ib)² = 1 ωi² It (It + Ib - It) = 1 ωi² ItIb 2 It + Ib 2 It + Ib
Correct Option: D
By conservation of angular momentum,
Itωi =(It + Ib) ωƒ
where ωƒ is the final angular velocity of disks∴ ωƒ = It ωi It + Ib
Loss in ΔK.E., = Initial K.E. – Final K.E.= 1 Itωi² - 1 (It + Ib)ωƒ² 2 2 ⇒ Δ K = 1 Itωi² - 1 (It + Ib) It² ωi² 2 2 (It + Ib)² = 1 ωi² It (It + Ib - It) = 1 ωi² ItIb 2 It + Ib 2 It + Ib
- The instantaneous angular position of a point on a rotating wheel is given by the equation θ(t) = 2t³ – 6t². The torque on the wheel becomes zero at
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When angular acceleration (α) is zero then torque on the wheel becomes zero.
θ(t) = 2t³ – 6t²⇒ dθ = 6t² - 12t ⇒ α = d²θ = 12t - 12 = 0 dt dt
∴ t = 1 sec.Correct Option: A
When angular acceleration (α) is zero then torque on the wheel becomes zero.
θ(t) = 2t³ – 6t²⇒ dθ = 6t² - 12t ⇒ α = d²θ = 12t - 12 = 0 dt dt
∴ t = 1 sec.
- If →F is the force acting on a particle having position vector →r and →τ be the torque of this force about the origin, then:
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Since, →τ is perpendicular to the plane of r' and, hence the dot product of →τ with r' and →F is zero.Correct Option: B
Since, →τ is perpendicular to the plane of r' and, hence the dot product of →τ with r' and →F is zero.
- A rod PQ of mass M and length L is hinged at end P. The rod is kept horizontal by a massless string tied to point Q as shown in figure. When string is cut, the initial angular acceleration of the rod is
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Weight of the rod will produce the torqueτ = mg L = Iα = mL² α 
∵ I rod = mL² 
2 3 3
Hence, angular acceleration α = 3g/2LCorrect Option: D
Weight of the rod will produce the torqueτ = mg L = Iα = mL² α ∵ I rod = mL² 2 3 3
Hence, angular acceleration α = 3g/2L
