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Rotational Motion

NEET Physics · 99 questions · Page 3 of 10 · Click an option or "Show Solution" to reveal answer

Q21
A solid sphere is rotating freely about its symmetry axis in free space. The radius of the sphere is increased keeping its mass same. Which of the following physical quantities would remain constant for the sphere?
A Angular velocity
B Moment of inertia
C Rotational kinetic energy
D Angular momentum
Correct Answer
Option D
Solution

As sphere is in free space and no external torque is acting over it so its angular momentum will remain constant.

Q22
A solid sphere is in rolling motion. In rolling motion a body possesses translational kinetic energy (K t ) as well as rotational kinetic energy (K r ) simultaneously. The ratio K t : (K t + K r ) for the sphere is
A 7 : 10
B 5 : 7
C 10 : 7
D 2 : 5
Correct Answer
Option B
Solution

In rolling motion, Translational kinetic energy, K t =

12mv2{1 \over 2}m{v^2}

Rotational kinetic energy, K r =

12Iω2{1 \over 2}I{\omega ^2}

∴\therefore K t + K r =

12mv2{1 \over 2}m{v^2}

+

12Iω2{1 \over 2}I{\omega ^2}

=

12mv2+12(25mr2)(vr)2{1 \over 2}m{v^2} + {1 \over 2}\left( {{2 \over 5}m{r^2}} \right){\left( {{v \over r}} \right)^2}

=

710mv2{7 \over {10}}m{v^2}

∴\therefore

KtKt+Kr=12mv2710mv2{{{K_t}} \over {{K_t} + {K_r}}} = {{{1 \over 2}m{v^2}} \over {{7 \over {10}}m{v^2}}}

=

57{5 \over 7}
Q23
The moment of the force, F→=4i^+5j^−6k^\overrightarrow F = 4\widehat i + 5\widehat j - 6\widehat k at (2, 0, –3), about the point (2, –2, –2), is given by
A −8i^−4j^−7k^ - 8\widehat i - 4\widehat j - 7\widehat k
B −4i^−j^−8k^ - 4\widehat i - \widehat j - 8\widehat k
C −7i^−8j^−4k^ - 7\widehat i - 8\widehat j - 4\widehat k
D −7i^−4j^−8k^ - 7\widehat i - 4\widehat j - 8\widehat k
Correct Answer
Option D
Solution

Given,

F→=4i^+5j^−6k^\overrightarrow F = 4\widehat i + 5\widehat j - 6\widehat k

We know, Moment of force,

τ→=(r→−r→0)×F→\overrightarrow \tau = \left( {\overrightarrow r - {{\overrightarrow r }_0}} \right) \times \overrightarrow F

Where

r→0=2i^−2j^−2k^{\overrightarrow r _0} = 2\widehat i - 2\widehat j - 2\widehat k

and

r→=2i^+0j^−3k^\overrightarrow r = 2\widehat i + 0\widehat j - 3\widehat k
r→−r→0{\overrightarrow r - {{\overrightarrow r }_0}}

=

(2i^+0j^−3k^)−(2i^−2j^−2k^)\left( {2\widehat i + 0\widehat j - 3\widehat k} \right) - \left( {2\widehat i - 2\widehat j - 2\widehat k} \right)

=

0i^+2j^−k^{0\widehat i + 2\widehat j - \widehat k}
τ→=(r→−r→0)×F→\overrightarrow \tau = \left( {\overrightarrow r - {{\overrightarrow r }_0}} \right) \times \overrightarrow F

=

∣i^j^k^02−145−6∣\left| \begin{array}{lll}{\widehat i} & {\widehat j} & {\widehat k} \\ 0 & 2 & { - 1} \\ 4 & 5 & { - 6} \end{array} \right|

= (-12 + 5)

i^{\widehat i}

- (+4)

j^{\widehat j}

+ (-8)

k^{\widehat k}

=

−7i^−4j^−8k^- 7\widehat i - 4\widehat j - 8\widehat k
Q24
Three objects, A : (a solid sphere), B : (a thin circular disk) and C : (a circular ring), each have the same mass M and radius R. They all spin with the same angular speed ω\omega about their own symmetry axes. The amounts of work (W) required to bring them to rest, would satisfy the relation
A W C > W B > W A
B W A > W B > W C
C W B > W A > W C
D W A > W C > W B
Correct Answer
Option A
Solution

Work done required to bring them rest

Δ\Delta

W =

Δ\Delta

K E (work-energy theorem)

Δ\Delta

W =

12Iω2{1 \over 2}I{\omega ^2}

For same ω\omega,

Δ\Delta

W ∞\infty I For a solid sphere, I A =

25MR2{2 \over 5}M{R^2}

For a thin circular disk, I B =

12MR2{1 \over 2}M{R^2}

For a circular ring, I C = MR 2 ∴\therefore I C > I B > I A So, W C > W B > W A

Q25
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?
A 0.25 rad s −-2
B 25 rad s −-2
C 5 m s −-2
D 25 m s −-2
Correct Answer
Option B
Solution

Given, mass of cylinder m = 3kg R = 40 cm = 0.4 m F = 30 N ; α\alpha = ?

As we know, torque τ\tau = Ia F × R = MR 2 α\alpha α\alpha =

F×RMR2{{F \times R} \over {M{R^2}}}
α=30×(0.4)3×(0.4)2\alpha = {{30 \times \left( {0.4} \right)} \over {3 \times {{\left( {0.4} \right)}^2}}}
⇒α=25 rad/s2\Rightarrow \alpha = 25\,rad/{s^2}
Q26
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{\omega _1} and ω2{\omega _2}. They are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is
A 14I(ω1−ω2)2{1 \over 4}I{\left( {{\omega _1} - {\omega _2}} \right)^2}
B I(ω1−ω2)2I{\left( {{\omega _1} - {\omega _2}} \right)^2}
C 18I(ω1−ω2)2{1 \over 8}I{\left( {{\omega _1} - {\omega _2}} \right)^2}
D 12I(ω1+ω2)2{1 \over 2}I{\left( {{\omega _1} + {\omega _2}} \right)^2}
Correct Answer
Option A
Solution

According to the problem, I

ω1{\omega _1}

+ I

ω2{\omega _2}

= 2I

ω0{\omega _0}
ω0{\omega _0}

=

(ω1−ω2)2{{\left( {{\omega _1} - {\omega _2}} \right)} \over 2}
Ki=12I(ω12+ω22){K_i} = {1 \over 2}I\left( {{\omega _1}^2 + {\omega _2}^2} \right)
Kf=14(ω1+ω2)2{K_f} = {1 \over 4}{\left( {{\omega _1} + {\omega _2}} \right)^2}

Loss

ΔK=I[ω122+ω222−ω124−ω224−2ω1ω24]\Delta K = I\left[ {{{{\omega _1}^2} \over 2} + {{{\omega _2}^2} \over 2} - {{{\omega _1}^2} \over 4} - {{{\omega _2}^2} \over 4} - {{2{\omega _1}{\omega _2}} \over 4}} \right]
=I[ω124+ω224−2ω1ω24]= I\left[ {{{{\omega _1}^2} \over 4} + {{{\omega _2}^2} \over 4} - {{2{\omega _1}{\omega _2}} \over 4}} \right]
=I4[ω1−ω2]2= {I \over 4}{\left[ {{\omega _1} - {\omega _2}} \right]^2}
Q27
A light rod of length ll has two masses m 1 and m 2 attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is
A m1m2m1+m2l2{{{m_1}{m_2}} \over {{m_1} + {m_2}}}{l^2}
B m1+m2m1m2l2{{{m_1} + {m_2}} \over {{m_1}{m_2}}}{l^2}
C (m1+m2)l2\left( {{m_1} + {m_2}} \right){l^2}
D m1m2 l2\sqrt {{m_1}{m_2}} \,{l^2}
Correct Answer
Option A
Solution

Here,

l1+l2=l{l_1} + {l_2} = l

Centre of mass of the system,

l1=m1×0+m2×lm1+m2=m2lm1+m2{l_1} = {{{m_1} \times 0 + {m_2} \times l} \over {{m_1} + {m_2}}} = {{{m_2}l} \over {{m_1} + {m_2}}}
l2=l−l1=m1lm1+m2{l_2} = l - {l_1} = {{{m_1}l} \over {{m_1} + {m_2}}}

Required moment of inertia of the system,

I=m1l12+m2l22I = {m_1}l_1^2 + {m_2}l_2^2
=(m1m22+m2m12)l2(m1+m2)2= \left( {{m_1}m_2^2 + {m_2}m_1^2} \right){{{l^2}} \over {{{\left( {{m_1} + {m_2}} \right)}^2}}}
=m1m2(m1+m2)l2(m1+m2)2=m1m2m1+m2l2= {{{m_1}{m_2}\left( {{m_1} + {m_2}} \right){l^2}} \over {{{\left( {{m_1} + {m_2}} \right)}^2}}} = {{{m_1}{m_2}} \over {{m_1} + {m_2}}}{l^2}
Q28
Two rotating bodies A and B of masses m and 2m with moments of inertia IA{I_A} and IB{I_B} (IB{I_B} > IA{I_A}) have equal kinetic energy of rotation. If L A and L B be their angular momenta respectively, then
A LA=LB2{L_A} = {{{L_B}} \over 2}
B LA=2LB{L_A} = 2{L_B}
C LB>LA{L_B} > {L_A}
D LA>LB{L_A} > {L_B}
Correct Answer
Option C
Solution

It is known that rotational kinetic energy is given as :

E=Iω22E = {{I{\omega ^2}} \over 2}

=

L22I{{{L^2}} \over {2I}}

Hence, L 2 ∝\propto I Further, E A = E B Also,

LA22IA=LB22IB{{{L_A}^2} \over {2{I_A}}} = {{{L_B}^2} \over {2{I_B}}}
IB>IA{I_B} > {I_A}
LB>LA{L_B} > {L_A}
Q29
A solid sphere of mass m and radius R is rotating about its diameter. A solid cylinder of same mass and same radius is also rotating about its geometrical axis with an angular speed twice that of the sphere. The ratio of their kinetic energies of rotation (E sphere / E cylinder ) will be
A 2 : 3
B 1 : 5
C 1 : 4
D 3 : 1
Correct Answer
Option B
Solution
ESphereECylinder=12Isωs212Icωc2=Isωs2Icωc2{{{E_{Sphere}}} \over {{E_{Cylinder}}}} = {{{1 \over 2}{I_s}\omega _s^2} \over {{1 \over 2}{I_c}\omega _c^2}} = {{{I_s}\omega _s^2} \over {{I_c}\omega _c^2}}

Here,

Is=25mR2{I_s} = {2 \over 5}m{R^2}

,

Ic=12mR2{I_c} = {1 \over 2}m{R^2}
ωc=2ωs{\omega _c} = 2{\omega _s}
ESphereECylinder=25mR2×ωs212mR2×(2ω2)2{{{E_{Sphere}}} \over {{E_{Cylinder}}}} = {{{2 \over 5}m{R^2} \times \omega _s^2} \over {{1 \over 2}m{R^2} \times {{\left( {2{\omega _2}} \right)}^2}}}
⇒45×14=15\Rightarrow {4 \over 5} \times {1 \over 4} = {1 \over 5}
Q30
A disc and a sphere of same radius but different masses roll off on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first?
A Both reach at the same time
B Depends on their masses
C Disc
D Sphere
Correct Answer
Option D
Solution

Time taken by the body to reach the bottom when it rolls down on an inclined plane without slipping is given by

t=2l(1+k2R2)gsin⁡θt = \sqrt {{{2l\left( {1 + {{{k^2}} \over {{R^2}}}} \right)} \over {g\sin \theta }}}

Since g is constant and

ll

, R and sin θ\theta are same for both ∴\therefore

tdts=1+kd2R21+ks2R2=1+R22R21+2R25R2{{{t_d}} \over {{t_s}}} = {{\sqrt {1 + {{k_d^2} \over {{R^2}}}} } \over {\sqrt {1 + {{k_s^2} \over {{R^2}}}} }} = \sqrt {{{1 + {{{R^2}} \over {2{R^2}}}} \over {1 + {{2{R^2}} \over {5{R^2}}}}}}
(∵kd=R2,ks=25R)\left(\because {{k_d} = {R \over {\sqrt 2 }},{k_s} = \sqrt {{2 \over 5}} R} \right)
⇒32×57=1514⇒td>ts\Rightarrow \sqrt {{3 \over 2} \times {5 \over 7}} = \sqrt {{{15} \over {14}}} \Rightarrow {t_d} > {t_s}
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