System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 25

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Question 1
1 / -0

A circular ring is rotating about an axis passing through its diameter. If its radius of gyration is $10 cm$ , diameter of the ring is:

A
$10\sqrt{2} cm$

B
$20 cm$

C
$20\sqrt{2} cm$

D
$40 cm$

Solution

By perpendicular axis theorem, moment of inertia about any two perpendicular axis is sum of moment of inertia's about any two areas in the plane
           $I=I_1+I_2$
       $2(I_1)=MR^2$ for a ring.
       $I_1=\dfrac{MR^2}{2}=M\left ( \dfrac{R}{\sqrt{2}} \right )^2$
with radius of gyration $\dfrac{R}{\sqrt{2}}=10$
       $\Rightarrow R=10\sqrt{2}$
$\therefore diameter=20\sqrt{2}$
Question 2
1 / -0

A circular disc of radius $10 \ cm$ is free to rotate about an axis passing through its centre without friction and its moment of inertia is $\dfrac{1}{2}\pi \ kg\ m^{2}$ . A tangential force of $20 N$ is acting on the disc along its rim. Starting from rest, the number of rotations made by the disc in $10 \ s$ is:

A
$50$

B
$100$

C
$150$

D
$200$

Solution

Torque on the disc $\tau ={F}\times{R}=20\times0.1=2Nm$
Given moment of inertia $I=\dfrac{1}{2\pi }kgm^2$
   $\tau =I\alpha$
$\Rightarrow \alpha =4\pi rads^{-2}$
No. of rotations made $=\dfrac{\left (\dfrac{1}{2}\:\alpha\: t^2 \right )}{2\pi }$ (by the 1st law of uniform angular rotation)

                                  $=\dfrac{\dfrac{1}{2}\times 4\pi \times 100 }{2\pi }$

                                $=100$
Question 3
1 / -0

The moment of inertia of thin rod is $2\times 10^{-3}kgm^{2}$ when it is made to rotate about an axis perpendicular to the length of rod and passing through one end. If half length of the rod is cut and removed then the moment of inertia of remaining rod is:

A
$10^{-3}kgm^{2}$

B
$0.5\times 10^{-3}kgm^{2}$

C
$0.25\times 10^{-3}kgm^{2}$

D
$25\times 10^{-3}kgm^{2}$

Solution

Moment of Inertia about one end $=\dfrac{ML^2}{3}$
When half the rod is cut $M'=\dfrac{M}{2}$ and $L'=\dfrac{L}{2}$

$\displaystyle\therefore I=\frac{1}{3}\frac{M}{2}\left (\frac{L}{2}\right )^2=\frac{ML^2}{3(8)}=\frac{I}{8}=\frac{2\times 10^{-3}}{8}kgm^2$

$=0.25\times 10^{-3}kgm^2$
Question 4
1 / -0

A circular disc is rotating about its own natural axis. A constant opposing torque $2.75 N m$ is applied on the disc due to which it comes to rest in $28$ rotations. If moment of inertia of disc is $0.5 kgm^{2}$ , the initial angular velocity of disc is:

A
$210 rpm$

B
$280 rpm$

C
$360 rpm$

D
$420 rpm$

Solution

we know that $\tau =I\alpha$
                $\Rightarrow \alpha =\dfrac{ \tau }{I}=\dfrac{2.75Nm}{0.5}=5.5\:rads^{-2}$
given no of rotations $=28$
             $\therefore Total\:angle\:\theta =28(2\pi )=56\pi$
$\omega^2=2\alpha \theta$
$\Rightarrow \omega^2=2\times 5.5 \times 56\pi$
        $\omega=43.99\:rad/s$
              $=43.99\times \dfrac{60}{2\pi }rpm=420\;rpm$
Question 5
1 / -0

A wheel of radius $10 cm$ can rotate freely about an axis passing through its centre. A force $20 N$ acts tangentially at the rim of wheel. If moment of inertia of wheel is $0.5 kgm^{2}$ , its angular acceleration is:

A
$2 \ rad/s^{2}$

B
$2.5 \ rad/s^{2}$

C
$4 \ rad/s^{2}$

D
$5 \ rad/s^{2}$

Solution

Torque on the wheel $\tau =\vec{F}\times \vec{R}$
as F is applied at rim.
$\Rightarrow \tau =20\times 0.1=2Nm$
Moment of Inertia $I=0.5kgm^2$
$\tau =I\alpha$
$\Rightarrow \alpha=\dfrac{\tau }{I}=\dfrac{2}{0.5}=4rad/s^2$
Question 6
1 / -0

A ceiling fan is rotating at the rate of $3.5 rps$ and its moment of inertia is $1.25 kgm^{2}$ . If the current is switched off, the fan comes to rest in $5.5 s$ . The torque acting on the fan due to friction is:

A
$2.5 \ Nm$

B
$5 \ N m$

C
$7.5 \ N m$

D
$10 \ N m$

Solution

$\tau =I\alpha$
$\alpha =\dfrac{d\omega }{dt}=\dfrac{3.5\times 2\pi }{5.5}=4$
$\therefore \tau =I\alpha =1.25\times4=5 \ Nm$
Question 7
1 / -0

The radius of gyration of rod of length $L$ and mass $M$ about an axis perpendicular to its length and passing through a point at a distance $\dfrac{L}{3}$ from one of its ends is:

A
$\dfrac{\sqrt{7}}{6}L$

B
$\dfrac{L}{9}$

C
$\dfrac{L}{3}$

D
$\dfrac{\sqrt{5}}{2}L$

Solution

Moment of inertia of rod about cm , $Icm=\dfrac{ML^2}{12}$
Distance of point from cm d= $l/2-l/3=l/6$ ,
So, Moment of inertia at that point $I=Icm+Md^2=\dfrac{ML^2}{12}+\dfrac{ML^2}{36}=\dfrac{ML^2}{9}$
So $ML^2/9=MK^2\Rightarrow K=L/3$
Question 8
1 / -0

Identify the increasing order of the radius of gyration of following bodies of same radius:
I) About natural axis of circular ring.
II) About diameter of circular ring.
III) About diameter of circular plate.
IV) About diameter of solid sphere.

A
II, III, IV, I

B
III, II, IV, I

C
III, IV, II, I

D
II, IV, III, I

Solution

(1) about Neutral axis of circular ring $\to$
$MR^{2}MRg^{2}$
$Rg=R$

(2) about diameter of circular ring $\to$
$\dfrac{MR^{2}}{2}=MRg^{2}$

$Rg=\dfrac{R}{\sqrt{2}}=0.707 R$

(3) about diameter of circular plate $\to$
$\dfrac{MR^{2}}{4}=MRg^{2}$

$Rg=\dfrac{R}{2}=0.5 R$

(4) about diameter of solid sphere $\to$
$\dfrac{MR^{2}}{5}=MRg^{2}$

$Rg=\sqrt{\dfrac{2}{5}}R=0.63 R$

Ans. $(III)<(IV)<(II)<(I)$
Question 9
1 / -0

A constant torque of $1000 N m$ turns a wheel of moment of inertia $200 kgm^{2}$ about an axis through its centre. The wheel is at rest initially. Its angular velocity after $3 s$ is :

A
$15 \ rads^{-1}$

B
$0.6 \ rads^{-1}$

C
$6000 \ rads^{-1}$

D
$5 \ rads^{-1}$

Solution

We know that $T=I\alpha$
$1000=200\times \alpha\Rightarrow \alpha =5rad/s^2$
now we know $\omega=\omega_o+\alpha t$ , and $\omega_o=0$
$\omega=5\times 3=15\ rads^{-1}$
Question 10
1 / -0

Identify the decreasing order of moments of inertia of the following bodies of same mass and radius.
I) about diameter of circular ring
II) about diameter of circular plate
III) about tangent of circular ring perpendicular to its plane
IV) about tangent of circular plate in its plane

A
III, IV, II, I

B
IV, III, I, II

C
IV, III, II, I

D
III, IV, I, II

Solution

Moment of inertia about diameter of circular ring $I_1=\dfrac{mr^2}{2}$
Moment of inertia about diameter of circular plate $I_2=\dfrac{mr^2}{4}$
Moment of inertia about tangent of circular ring perpendicular to its plane $I_3=mr^2+mr^2=2mr^2$
Moment of inertia about tangent of circular plate in its plane $I_4=mr^2/4+mr^2=\dfrac{5mr^2}{4}$
So $I_3>I_4>I_1>I_2$

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System of Particles and Rotational Motion Test - 25

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System of Particles and Rotational Motion Test - 25

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Question 1

102 cm10\sqrt{2} cm102​ cm

20 cm20 cm20 cm

202 cm20\sqrt{2} cm202​ cm

40 cm40 cm40 cm

Solution

Question 2

505050

100100100

150150150

200200200

Solution

Question 3

10−3kgm210^{-3}kgm^{2}10−3kgm2

0.5×10−3kgm20.5\times 10^{-3}kgm^{2}0.5×10−3kgm2

0.25×10−3kgm20.25\times 10^{-3}kgm^{2}0.25×10−3kgm2

25×10−3kgm225\times 10^{-3}kgm^{2}25×10−3kgm2

Solution

Question 4

210 rpm210 rpm210 rpm

280 rpm280 rpm280 rpm

360 rpm360 rpm360 rpm

420 rpm420 rpm420 rpm

Solution

Question 5

2 rad/s22 \ rad/s^{2}2 rad/s2

2.5 rad/s22.5 \ rad/s^{2}2.5 rad/s2

4 rad/s24 \ rad/s^{2}4 rad/s2

5 rad/s25 \ rad/s^{2}5 rad/s2

Solution

Question 6

2.5 Nm2.5 \ Nm2.5 Nm

5 Nm5 \ N m5 Nm

7.5 Nm7.5 \ N m7.5 Nm

10 Nm10 \ N m10 Nm

Solution

Question 7

76L\dfrac{\sqrt{7}}{6}L67​​L

L9\dfrac{L}{9}9L​

L3\dfrac{L}{3}3L​

52L\dfrac{\sqrt{5}}{2}L25​​L

Solution

Question 8

II, III, IV, I

III, II, IV, I

III, IV, II, I

II, IV, III, I

Solution

Question 9

15 rads−115 \ rads^{-1}15 rads−1

0.6 rads−10.6 \ rads^{-1}0.6 rads−1

6000 rads−16000 \ rads^{-1}6000 rads−1

5 rads−15 \ rads^{-1}5 rads−1

Solution

Question 10

III, IV, II, I

IV, III, I, II

IV, III, II, I

III, IV, I, II

Solution

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$10\sqrt{2} cm$

$20 cm$

$20\sqrt{2} cm$

$40 cm$

$50$

$100$

$150$

$200$

$10^{-3}kgm^{2}$

$0.5\times 10^{-3}kgm^{2}$

$0.25\times 10^{-3}kgm^{2}$

$25\times 10^{-3}kgm^{2}$

$210 rpm$

$280 rpm$

$360 rpm$

$420 rpm$

$2 \ rad/s^{2}$

$2.5 \ rad/s^{2}$

$4 \ rad/s^{2}$

$5 \ rad/s^{2}$

$2.5 \ Nm$

$5 \ N m$

$7.5 \ N m$

$10 \ N m$

$\dfrac{\sqrt{7}}{6}L$

$\dfrac{L}{9}$

$\dfrac{L}{3}$

$\dfrac{\sqrt{5}}{2}L$

$15 \ rads^{-1}$

$0.6 \ rads^{-1}$

$6000 \ rads^{-1}$

$5 \ rads^{-1}$