System of Particles and Rotational Motion
- From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing
through the centre ?
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Moment of inertia of complete disc about point 'O'.
ITotal disc = MR²/2
Mass of removed disc
MRemoved = M/4 (Mass ∝ area)
Moment of inertia of removed disc about point 'O'.
IRemoved (about same perpendicular axis)
= Icm + mx²= M (R/2)² + M 
R 
² = 3MR² 4 2 4 2 32
Therefore the moment of inertia of the remaining part of the disc about a perpendicular axis passing through the centre,
IRemaing disc = ITotal – IRemoved= MR² = 3 MR² = 13 MR² 2 32 32 Correct Option: B
Moment of inertia of complete disc about point 'O'.
ITotal disc = MR²/2
Mass of removed disc
MRemoved = M/4 (Mass ∝ area)
Moment of inertia of removed disc about point 'O'.
IRemoved (about same perpendicular axis)
= Icm + mx²= M (R/2)² + M R ² = 3MR² 4 2 4 2 32
Therefore the moment of inertia of the remaining part of the disc about a perpendicular axis passing through the centre,
IRemaing disc = ITotal – IRemoved= MR² = 3 MR² = 13 MR² 2 32 32
- Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities ω1 and ω2 . They are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is:-
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Here, lω1 + lω2 = 2lω
⇒ ω = ω1 + ω2 2 (K.E)i = 1 lω1² + 1 lω2² 2 2 (K.E)f = 1 2lω² = T ω1 + ω2 ² 2 2
Loss in K.E. = (K.E)f - (K.E)i = (1/4) l(ω1 - ω2)²Correct Option: A
Here, lω1 + lω2 = 2lω
⇒ ω = ω1 + ω2 2 (K.E)i = 1 lω1² + 1 lω2² 2 2 (K.E)f = 1 2lω² = T ω1 + ω2 ² 2 2
Loss in K.E. = (K.E)f - (K.E)i = (1/4) l(ω1 - ω2)²
- A particle of mass m = 5 is moving with a uniform speed v = 3√2 in the XOY plane along the line y = x + 4. The magnitude of the angular momentum of the particle about the origin is
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y = x + 4 line has been shown in the figure when x = 0, y = 4. So, OP = 4.
The slope of the line can be obtained by comparing with the equation of the straight line
y = mx + c
m = tan θ = 1
⇒ θ = 45°
∠OQP = ∠OPQ = 45°
If we draw a line perpendicular to this line,
length of the perpendicular = OR
In ∆OPR, OR/OP = sin 45°
⇒ OR = OP sin 45°= 4 × 1 = 4 = 2√2 √2 √2
Angular momentum of particle going along
this line = r × mv = 2√2 × 5 × 3√2 = 60 units
Correct Option: A
y = x + 4 line has been shown in the figure when x = 0, y = 4. So, OP = 4.
The slope of the line can be obtained by comparing with the equation of the straight line
y = mx + c
m = tan θ = 1
⇒ θ = 45°
∠OQP = ∠OPQ = 45°
If we draw a line perpendicular to this line,
length of the perpendicular = OR
In ∆OPR, OR/OP = sin 45°
⇒ OR = OP sin 45°= 4 × 1 = 4 = 2√2 √2 √2
Angular momentum of particle going along
this line = r × mv = 2√2 × 5 × 3√2 = 60 units
- A constant torque of 1000 N-m turns a wheel of moment of inertia 200 kg-m² about an axis through its centre. Its angular velocity after 3 seconds is
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τ = 1000 N - m, I = 200 kg-m²
∴ Iα = 1000
⇒ α = 1000/200 = 5 rad/sec²
ω = ω0 + αt = 0 + 3 × 5 = 15 rad/sCorrect Option: D
τ = 1000 N - m, I = 200 kg-m²
∴ Iα = 1000
⇒ α = 1000/200 = 5 rad/sec²
ω = ω0 + αt = 0 + 3 × 5 = 15 rad/s
- A round disc of moment of inertia I2 about its axis perpendicular to its plane and passing through its centre is placed over another disc of moment of inertia I1 rotating with an angular velocity ω about the same axis. The final angular velocity of the combination of discs is
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Angular momentum will be conserved
I1ω = I1ω' + I2ω' ⇒ ω' = I1ω l1 + l2 Correct Option: D
Angular momentum will be conserved
I1ω = I1ω' + I2ω' ⇒ ω' = I1ω l1 + l2
