Practice Start Test

Rotational Motion

NEET Physics · 99 questions · Page 6 of 10 · Click an option or "Show Solution" to reveal answer

Q51
A small mass attached to a string rotates on a frictionless table top as shown. If the tension in the string is increased by pulling the string causing the radius of the circular motion to decrease by a factor of 2, the kinetic energy of the mass will
A decrease by a factor of 2
B remain constant
C increase by a factor of 2
D increase by a factor of 4
Correct Answer
Option D
Solution

According to law of conservation of angular momentum mvr = mv'r'

vr=v′(r2)vr = v'\left( {{r \over 2}} \right)
v′=2vv' = 2v

...(i) ∴\therefore

KK′=12mv212mv′2=(vv′)2{K \over {K'}} = {{{1 \over 2}m{v^2}} \over {{1 \over 2}mv{'^2}}} = {\left( {{v \over {v'}}} \right)^2}
⇒K′K=(v′v)2=(2)2\Rightarrow {{K'} \over K} = {\left( {{{v'} \over v}} \right)^2} = {\left( 2 \right)^2}

(Using (i))

⇒K′=4K\Rightarrow K' = 4K
Q52
The instantaneous angular position of a point on a rotating wheel is given by the equation θ(t)=2t3−6t2\theta \left( t \right) = 2{t^3} - 6{t^2} The torque on the wheel becomes zero at
A t = 1 s
B t = 0.5 s
C t = 0.25 s
D t = 2 s
Correct Answer
Option A
Solution

When angular acceleration (a) is zero then torque on the wheel becomes zero.

θ(t)=2t3−6t2\theta \left( t \right) = 2{t^3} - 6{t^2}
⇒dθdt=6t2−12t\Rightarrow {{d\theta } \over {dt}} = 6{t^2} - 12t
⇒α=d2θdt2=12t−12=0\Rightarrow \alpha = {{{d^2}\theta } \over {d{t^2}}} = 12t - 12 = 0

∴\therefore t = 1 sec.

Q53
The moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its midpoint and perpendicular to its length is II 0 . Its moment of inertia about an axis passing through one of its ends and perpendicular to its length is
A I0+ML2/2{I_0} + M{L^2}/2
B I0+ML2/4{I_0} + M{L^2}/4
C I0+2ML2{I_0} + 2M{L^2}
D I0+ML2{I_0} + M{L^2}
Correct Answer
Option B
Solution

By the theorem of parallel axes, the moment of inertia of a body about any axis is equal to the sum of the moment of inertia about a parallel axis through the centre of mass and Ma 2 , where M is the mass of the body and a is the separation between the two axes.

Therefore, the moment of inertia of the rod about an axis passing through one of its ends and perpendicular to its length is I = I 0 + M (L/2) 2 = I 0 + ML 2 /4

Q54
A solid cylinder and a hollow cylinder, both of the same mass and same external diameter are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first?
A Both together only when angle of inclination of plane is 45 o
B Both together
C Hollow cylinder
D Solid cylinder
Correct Answer
Option D
Solution

Time taken to reach the bottom of inclined plane.

t=2l(1+K2R2)gsin⁡θt = \sqrt {{{2l\left( {1 + {{{K^2}} \over {{R^2}}}} \right)} \over {g\sin \theta }}}

Here,

ll

is length of incline plane For solid cylinder

K2=R22{K^2} = {{{R^2}} \over 2}

For hollow cylinder K 2 = R 2 Hence, solid cylinder will reach the bottom first.

Q55
From a circular disc of radius R and mass 9M, a small disc of mass M and radius R3{R \over 3} is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is
A 409{{40} \over 9} MR 2
B MR 2
C 4MR 2
D 49{4 \over 9} MR 2
Correct Answer
Option A
Solution

Mass of the disc = 9M Mass of removed portion of disc = M The moment of inertia of the complete disc about an axis passing through its centre O and perpendicular to its plane is

I1=92MR2{I_1} = {9 \over 2}M{R^2}

Now, the moment of inertia of the disc with removed portion

I2=12M(R3)2=118MR2{I_2} = {1 \over 2}M{\left( {{R \over 3}} \right)^2} = {1 \over {18}}M{R^2}

Therefore, moment of inertia of the remaining portion of disc about O is

I=I1−I2=92MR2−MR218=40MR29I = {I_1} - {I_2} = {9 \over 2}M{R^2} - {{M{R^2}} \over {18}} = {{40M{R^2}} \over 9}
Q56
A thin circular ring of mass M and radius r is rotating about its axis with constant angular velocity ω\omega . Two objects each of msass m are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with angular velocity given by
A (M+2m)ω2m{{\left( {M + 2m} \right)\omega } \over {2m}}
B 2MωM+2m{{2M\omega } \over {M + 2m}}
C (M+2m)ωM{{\left( {M + 2m} \right)\omega } \over M}
D MωM+2m{{M\omega } \over {M + 2m}}
Correct Answer
Option D
Solution

As no external torque is acting about the axis, angular momentum of system remains conserved. I 1

ω1{\omega _1}

= I 2

ω2{\omega _2}
⇒ω2=I1ω1(M+2m)r2=Mω(M+2m)\Rightarrow {\omega _2} = {{{I_1}{\omega _1}} \over {\left( {M + 2m} \right){r^2}}} = {{M\omega } \over {\left( {M + 2m} \right)}}
Q57
A gramophone record is revolving with an angular velocity ω\omega . A coin is placed at a distance r from the centre of the record. The static coefficient of friction is μ\mu . The coin will revolve with the record if
A r = μ\mu gω\omega 2
B r < ω2μg{{{\omega ^2}} \over {\mu g}}
C r≤μgω2r \le {{\mu g} \over {{\omega ^2}}}
D r≥μgω2r \ge {{\mu g} \over {{\omega ^2}}}
Correct Answer
Option C
Solution

The coin will revolve with the record, if Force of friction ≥\ge Centrifugal force

μmg≥mrω2\mu mg \ge mr{\omega ^2}
⇒r≤μgω2\Rightarrow r \le {{\mu g} \over {{\omega ^2}}}
Q58
A circular disk of moment of inertia It{I_t} is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed ωi{\omega _i}. Another disk of moment of inertia Ib{I_b} 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 ω\omega . The energy lost by the initially rotating disc to friction is
A 12Ib2(It+Ib)ωi2{1 \over 2}{{I_b^2} \over {\left( {{I_t} + {I_b}} \right)}}\omega _i^2
B 12It2(It+Ib)ωi2{1 \over 2}{{I_t^2} \over {\left( {{I_t} + {I_b}} \right)}}\omega _i^2
C Ib−It(It+Ib)ωi2{{{I_b} - {I_t}} \over {\left( {{I_t} + {I_b}} \right)}}\omega _i^2
D 12IbIt(It+Ib)ωi2{1 \over 2}{{{I_b}{I_t}} \over {\left( {{I_t} + {I_b}} \right)}}\omega _i^2
Correct Answer
Option D
Solution

As no external torque is applied to the system, the angular momentum of the system remains conserved.

∴\therefore L i = L f According to given problem,

Itωi=(It+Ib)ωf{I_t}{\omega _i} = \left( {{I_t} + {I_b}} \right){\omega _f}
⇒ωf=Itωi(It+Ib)\Rightarrow {\omega _f} = {{{I_t}{\omega _i}} \over {\left( {{I_t} + {I_b}} \right)}}

...(i) Initial energy,

Ei=12Itωi2{E_i} = {1 \over 2}{I_t}\omega _i^2

...(ii) Final energy,

Ef=12(It+Ib)ωf2{E_f} = {1 \over 2}\left( {{I_t} + {I_b}} \right)\omega _f^2

...(iii) Substituting the value of

ωf{\omega _f}

from equation (i) in equation (iii), we get Final energy,

Ef=12(It+Ib)(Itωi(It+Ib))2{E_f} = {1 \over 2}\left( {{I_t} + {I_b}} \right){\left( {{{{I_t}{\omega _i}} \over {\left( {{I_t} + {I_b}} \right)}}} \right)^2}
=12It2ωi2It+Ib= {1 \over 2}{{I_t^2\omega _i^2} \over {{I_t} + {I_b}}}

...(iv) Loss of energy,

Δ\Delta

E = E i – E f

=12Itωi2−12It2ωi2(It+Ib)= {1 \over 2}{I_t}\omega _i^2 - {1 \over 2}{{I_t^2\omega _i^2} \over {\left( {{I_t} + {I_b}} \right)}}

(Using (ii) and (iv))

=ωi22(It−It2(It+Ib))=ωi22(It2−IbIt−It2(It+Ib))= {{\omega _i^2} \over 2}\left( {{I_t} - {{I_t^2} \over {\left( {{I_t} + {I_b}} \right)}}} \right) = {{\omega _i^2} \over 2}\left( {{{I_t^2 - {I_b}{I_t} - I_t^2} \over {\left( {{I_t} + {I_b}} \right)}}} \right)
=12IbIt(It+Ib)ωi2= {1 \over 2}{{{I_b}{I_t}} \over {\left( {{I_t} + {I_b}} \right)}}\omega _i^2
Q59
A thin circular ring of mass M and radius R is rotating in a horizontal plane about an axis vertical to its plane with a constant angular velocity ω\omega . If two objects each of mass m be attached gently to the opposite ends of a diameter of the ring, the ring will then rotate with an angular velocity
A ωMM+2m{{\omega M} \over {M + 2m}}
B ω(M+2m)M{{\omega \left( {M + 2m} \right)} \over M}
C ωMM+m{{\omega M} \over {M + m}}
D ω(M−2m)M+2m{{\omega \left( {M - 2m} \right)} \over {M + 2m}}
Correct Answer
Option A
Solution

As the masses are added to the ring gently, there is no torque and angular momentum is conserved.

Iω=I′ω′I\omega = I'\omega '
⇒MR2ω=(MR2+2mR2)ω′\Rightarrow M{R^2}\omega = \left( {M{R^2} + 2m{R^2}} \right)\omega '
⇒ω′=MR2ω(M+2m)R2\Rightarrow \omega ' = {{M{R^2}\omega } \over {\left( {M + 2m} \right){R^2}}}
⇒ω′=MωM+2m\Rightarrow \omega ' = {{M\omega } \over {M + 2m}}
Q60
Four identical thin rods each of mass M and length ll, form a square frame. Moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane is
A 23Ml2{2 \over 3}M{l^2}
B 133Ml2{{13} \over 3}M{l^2}
C 13Ml2{1 \over 3}M{l^2}
D 43Ml2{4 \over 3}M{l^2}
Correct Answer
Option D
Solution

As per theorem of parallel axes : I = I cm + Ma 2 where I cm = moment of inertia about an axis through centre of mass of rod I = moment of inertia about parallel axis at distance ‘a’ Now moment of inertia of thin rod of mass M and length L about an axis through the midpoint of the rod and perpendicular to its length is ML 2 /12.

The moment of inertia of a rod about parallel axis at distance L/2 will be : ML 2 /12 + M(L/2) 2 = ML 2 /3 As there are 4 rods in square frame, so moment of inertia of entire frame is 4 times the moment of inertia of 1 rod, so it will be (4/3) ML 2

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