System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 65

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Question 1
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Directions For Questions

The figure shows an isosceles triangular plate of mass $$M$$ and base $$L$$. The angle at the apex is $$90^{\small\circ}$$.
the apex lies at the origin and the base is parallel to y-axis.

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The moment of inertia of the plate about z-axis is

A
$$\displaystyle\frac{ML^2}{12}$$

B
$$\displaystyle\frac{ML^2}{24}$$

C
$$\displaystyle\frac{ML^2}{6}$$

D
None of these

Solution

By unfolding, we can treat the given scenario as the square sheet of side $$L$$ rotating about O.
Now as MOI of rod of length $$a$$ about its centre of mass is $$\dfrac{Ma^2}{12}$$
Thus $$I_{x'} = I_{y'} = \dfrac{M(L)^2}{12} $$
Now using perpendicular axis theorem: $$I_z = I_{x'}+I_{y'}$$
$$I_z =2I_{x'} = \dfrac{ML^2}{6}$$
Question 2
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Two men '$$A$$' and '$$B$$' are standing on a plank. '$$B$$' is at the middle of the plank and '$$A$$' is at the left end of the plank. Surface of the plank is smooth. System is initially at rest and masses are as shown in figure. '$$A$$' and '$$B$$' start moving such that the position of '$$B$$' remains fixed with respect to ground and '$$A$$' meets '$$B$$'. Then the point where $$A$$ meets $$B$$ is located at

A
the middleof the plank

B
30cm from the left end of the. plank

C
the right end of the plank

D
None of these

Solution

Here in the given question it is mention that '$$B$$' start moving such that the position of '$$B$$' remains fixed with respect to ground, which clearly states that both planks $$ "A"$$ and "$$B$$" are moving with the same velocity.
Now it is clear from the observation that both will meet at right of "$$B$$";
Mathematically it can be expressed as:
Let velocity of "$$A$$"=$${v}_{1}$$
The velocity of "$$B$$" and plank=$${v}_{2}$$
Now $$S=$$($${{v}_{1}}-{{v}_{2}}){\times}t$$
$${v}_{1}$$=$$\cfrac{S}{t}+{{v}_{2}}$$
Hence it proves that $${v}_{1}$$ is greater than $${v}_{2}$$, and they both meet at the right end of the plank.
Question 3
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There are two particles of same mass. If one of the particles is at rest always and the other has an acceleration $$\bar{a}$$. Acceleration of center of mass is :

A
zero

B
$$\displaystyle\frac{1}{2}\, \bar{a}$$

C
$$\bar{a}$$

D
centre of mass for such a system can not be defined.

Solution

$${a}_{c.o.m}$$=$$\dfrac{{m}_{1}\times{a}_{1}+{m}_{2}\times{a}_{2}}{{m}_{1}+{m}_{2}}$$
given both body having same mass so
$${a}_{c.o.m}$$=$$\dfrac{{m}\times{a}_{1}+{m}\times{a}_{2}}{{2m}}$$
$${a}_{1}$$=0 given that body 1 is in rest
so $${\bar{a}}_{c.o.m}$$=$$\dfrac{1}{2}$$$$\times\bar{a}$$
Question 4
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A ceiling fan rotates about its own axis with some angular velocity. When the fan is switched off, the angular velocity becomes $$\left (\dfrac {1}{4}\right )^{th}$$ of the original in time $$'t'$$ and $$'n'$$ revolutions are made in that time. The number of revolutions made by the fan during the time interval between switch off and rest are (Angular retardation is uniform).

A
$$\dfrac {4n}{15}$$

B
$$\dfrac {8n}{15}$$

C
$$\dfrac {16n}{15}$$

D
$$\dfrac {32n}{15}$$

Solution

$$(\dfrac{\omega}{4})^2= \omega^2 - 2 \alpha \ n (2\pi)$$
$$\therefore 2 \alpha n (2 \pi) = \omega^2 - \dfrac{\omega^2}{16}$$
$$\therefore 2 \pi n = \dfrac{15}{16} \times (\dfrac{\omega^2}{2 \alpha})$$
$$0= \omega^2 - 2 \alpha n'$$
$$2 \pi n' = \dfrac{\omega^2}{2 \alpha }$$
$$\therefore n' = \dfrac{16n}{15}$$
Question 5
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If torques of equal magnitudes are applied to a hollow cylinder and a solid sphere both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry and the sphere is free to rotate about an axis passing through its center. Which of the two will acquire a greater angular speed after a given time?

A
$${ \omega }_{ 1 } > { \omega }_{ 2 }$$

B
$${ \omega }_{ 1 } = { \omega }_{ 2 }$$

C
$${ \omega }_{ 2 } > { \omega }_{ 1 }$$

D
None of these

Solution

Consider $${ I }_{ 1 }$$ and $${ I }_{ 2 }$$ be the moments of inertia of the hollow cylinder and solid sphere about its axis through its center respectively.
Then, $${ I }_{ 1 }=M{ R }^{ 2 }$$ ....(i)
and $${ I }_{ 2 }=\dfrac { 2 }{ 5 } M{ R }^{ 2 }$$ ....(ii)
Let $$\tau$$ be the magnitude of the torque applied on each of them. If $${ \alpha }_{ 1 }$$ and $${ \alpha }_{ 2 }$$ are the angular accelerations produced in the cylinder and sphere respectively, then
$$\tau ={ I }_{ 1 }{ \alpha }_{ 1 }$$
and $$\tau ={ I }_{ 2 }{ \alpha }_{ 2 }$$
$$\therefore { I }_{ 1 }{ \alpha }_{ 1 }={ I }_{ 2 }{ \alpha }_{ 2 }\Rightarrow \dfrac { { \alpha }_{ 1 } }{ { \alpha }_{ 2 } } =\dfrac { { I }_{ 2 } }{ { I }_{ 1 } }$$
$$=\dfrac { \dfrac { 2 }{ 5 } M{ R }^{ 2 } }{ M{ R }^{ 2 } } =\dfrac { 2 }{ 5 } \Rightarrow { \alpha }_{ 2 }=\dfrac { 5 }{ 2 } { \alpha }_{ 1 }$$
$$\Rightarrow { \alpha }_{ 2 }=2.5{ \alpha }_{ 1 }$$ ....(iii)
If $${ \omega }_{ 1 }$$ and $${ \omega }_{ 2 }$$ be the angular speed of the cylinder and sphere after time $$t$$, then
$${ \omega }_{ 1 }={ \omega }_{ 0 }+{ \alpha }_{ 1 }t$$ ....(iv)
and $${ \omega }_{ 2 }={ \omega }_{ 0 }+{ \alpha }_{ 2 }t$$
$$={ \omega }_{ 0 }+2.5{ \alpha }_{ 1 }t$$ .....(v)
where, $${ \omega }_{ 0 }=$$ initial angular speed
$$\therefore$$ From equation (iv) and (v), it is clear that
$$ { \omega }_{ 2 } > { \omega }_{ 1 }$$
$$\therefore $$ The sphere will acquire more angular speed as compared to that of the cylinder after a given time.
Question 6
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Fill in the blanks.
The effect of rotation of the earth is ......... at the equator and .............. at the poles. If the earth stops rotating. the weight of a body would .......... due to the absence of .............. force.

A
Minimum, maximum, decreases, gravitional

B
Maximum, minimum, increases, centrifugal

C
Maximum, minimum, decreases, centrifugal

D
Minimum, maximum, increases, gravitional

Solution
Question 7
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A bar magnet of moment of inertia $$I$$ is vibrated in a magnetic field of induction $$\displaystyle 0.4\times { 10 }^{ -4 }T$$. The time period of vibration is $$12\ s$$. The magnetic moment of the magnet is $$\displaystyle 120\ { Am }^{ 2 }$$. The moment of inertia of the magnet is (in $$\displaystyle kg{ m }^{ 2 }$$) approximately:

A
$$\displaystyle 2.1{ \pi }^{ 2 }\times{ 10 }^{ -2 }$$

B
$$\displaystyle 1.57\times { 10 }^{ -2 }$$

C
$$\displaystyle 1728\times { 10 }^{ -2 }$$

D
$$\displaystyle 172.8\times { 10 }^{ -4 }$$

Solution

An SHM takes place due to torque acting on magnet due to presence of magnetic field.
$$\tau=\mu Bsin\theta\approx \mu B\theta=I\omega^2\theta$$
$$\implies I=\cfrac{\mu B}{\omega ^2}=\cfrac{\mu BT^2}{4\pi ^2}$$
$$I=\cfrac{120 \times 0.4 \times 10^{-4} \times 12^2}{4 \times \pi^2}$$
$$=172.8\times 10^{-4} {kgm}^2$$
Question 8
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A uniform rod of length $$l=1m$$ is kept as shown in the figure. $$H$$ is a horizontal smooth surface and $$W$$ is a vertical smooth well. The rod is release from this position. What is the angular angular acceleration of the rod just after the released?

A
$$\dfrac {6g\cos \theta}{l}$$

B
$$\dfrac {3g\cos \theta}{l}$$

C
$$6g\cos \theta$$

D
$$\dfrac {2g\cos \theta}{l}$$

Solution

By balancing forces along $$H$$ and $$W$$.
$${ N }_{ 1 }=mg$$
$${ N }_{ 2 }=0$$
Torque about COM
$$\cfrac { l }{ 2 } \times { N }_{ 1 }\cos\theta \\ \cfrac { l }{ 2 } \times mg.\cfrac { 3 }{ 5 } =\cfrac { 3mgl }{ 10 } $$
About COM
$$\cfrac { l }{ 12 } \times m{ l }^{ 2 }\\ \therefore \alpha =\cfrac { z }{ I } =\cfrac { 3mgl }{ 10m{ l }^{ 2 } } \times 12=\cfrac { 36g }{ 10l } =35.28$$
Question 9
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A hoop of mass $$m$$ is projected on a floor with linear velocity $$v_{0}$$ and reverse spin $$\omega_{0}$$. The coefficient of friction between the hoop and the ground is $$\mu$$.
a. Under what condition will the hoop return back?
b. How far will it go?
c. How long will it continue to slip when its centre of mass becomes stationary?
d. What is the velocity of return?

A
a. $$\omega_{0} > \dfrac {v_{0}}{R}$$.
b. $$\dfrac {v_{0}^{2}}{R}$$,
c. $$\dfrac {R}{3\mu_{g}}\left (\omega_{0} - \dfrac {v_{0}}{R}\right )$$,
d. $$\dfrac {\omega_{0}R}{2} - \dfrac {v_{0}}{2}$$.

B
a. $$\omega_{0} > \dfrac {v_{0}}{R}$$.
b. $$\dfrac {v_{0}^{3}}{R}$$,
c. $$\dfrac {R}{2\mu_{g}}\left (\omega_{0} - \dfrac {v_{0}}{R}\right )$$,
d. $$\dfrac {\omega_{0}R}{4} - \dfrac {v_{0}}{2}$$.

C
a. $$\omega_{0} > \dfrac {v_{0}}{R}$$.
b. $$\dfrac {v_{0}^{2}}{R}$$,
c. $$\dfrac {R}{2\mu_{g}}\left (\omega_{0} - \dfrac {v_{0}}{R}\right )$$,
d. $$\dfrac {\omega_{0}R}{3} - \dfrac {v_{0}}{2}$$.

D
a. $$\omega_{0} > \dfrac {v_{0}}{R}$$.
b. $$\dfrac {v_{0}^{2}}{R}$$,
c. $$\dfrac {R}{2\mu_{g}}\left (\omega_{0} - \dfrac {v_{0}}{R}\right )$$,
d. $$\dfrac {\omega_{0}R}{2} - \dfrac {v_{0}}{2}$$.

Solution
Question 10
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Calculate the moment of inertia of a uniform rod of mass $$M$$ and length $$l$$ about an axis passing through an end and perpendicular to the rod. The rod can be divided into a number of mass elements along the length of the rod.

A
$$\cfrac { M{ l }^{ 2 } }{ 3 } $$

B
$$\cfrac { M{ l }^{ 2 } }{ 2 } $$

C
$$\cfrac { M{ l }^{ 2 } }{ 4 } $$

D
$$\cfrac { M{ l }^{ 3 } }{ 5 } $$

Solution

We have,
$$\Rightarrow dI=x^2xdm$$                 $$x-$$ distance from axis
                                       $$dm-$$ small mass element
$$dm-$$ mass of $$dx$$
$$\therefore dm=\dfrac{M}{l}dx$$
$$\therefore dI=x^2\dfrac{M}{l}dx$$
$$\therefore I=\int { \dfrac { M }{ l } { x }^{ 2 }.dx } =\dfrac { M }{ l } \int _{ 0 }^{ l }{ { x }^{ 2 }.dx } $$
        $$=\dfrac{M}{l}{ \left[ \dfrac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ l }$$
$$\therefore I=\dfrac{Ml^3}{3l}=\dfrac{Ml^2}{3}$$
Hence, the answer is $$\int\left(dm\right)r^2-\int _{ 0 }^{ l }\left( \dfrac { M }{ l } dx \right) { x }^{ 2 }=\dfrac { M{ l }^{ 2 } }{ 3 } $$