Tag: couple

Questions Related to couple

When slightly different weights are placed on the two pans of a beam balance, the beam comes to rest at an angle with the horizontal. The beam is supported at a single point P by a pivot. Then which of the following statement(s) is/are true ?

  1. The net torque about P due to the two weights is nonzero at the equilibrium position.

  2. The whole system does not continue to rotate about P because it has a large moment of inertia.

  3. The centre of mass of the system lies below P.

  4. The centre of mass of the system lies above P.


Correct Option: A
Explanation:

The whole system does not continue to rotate about P because the moment is balanced. Thus option B is wrong. And the center of mass of the system lies at pivot point P. Thus option C and D are wrong. As the force applied at the two points of suspension is different $\tau$ is different.

When a ceiling fan is switched off, its angular velocity reduces by $50$% while it makes $36$ rotations. How many more rotations will it make before coming to rest? (Assume uniforms angular retardation)

  1. $48$

  2. $36$

  3. $12$

  4. $18$


Correct Option: C
Explanation:

$\begin{array}{l} You\, \, have\, \, to\, \, use\, \, the\, \, equation, \ \; { { { \omega  } }^{ { 2 } } }\; ={ { { \omega  } } _{ { 0 } } }^{ { 2 } }\; +{ { 2\alpha \theta  } }\; \, \, for\, \, finding\, \, the\, \, angular\, \, acceleration\; \, \alpha \, \, and \ hence\, \, the\, \, number\, \, of\, \, further\, \, rotations. \ Note\, \, that\, \, this\, \, equation\, \, is\, \, the\, \, rotational\, \, analogue\, \, of\, \, the\, equation \ { v^{ 2 } }\; ={ v _{ 0 } }^{ 2 }+2as{ {  } }(or,\; { v^{ 2 } }\; ={ u^{ 2 } }\; +2as)\, \, in\, \, linear\, \, motion. \ Since\, \, the\, \, angular\, \, velocity\, \, has\, \, reduce\, \, to\, \, half\, \, of\, \, the\, \, initial\, \, value\, \, { \omega _{ 0 } }\, \, after\, \, 36\, \, rotations,\, \, we\, \, have \ { \left( { { \omega _{ 0\;  } }/2 } \right) _{ \;  } }^{ 2 }={ \omega _{ 0 } }^{ 2 }+2\alpha \times 36\, \, from\, \, which\, \; \alpha =--\; { \omega _{ 0 } }^{ 2 }/96 \ \left[ { We\, \, have\, \, expressed\, \, the\, \, angular\, \, displacement\, \, \theta \, \, in\, \, rotations\, \, itself\, \, for\, \, convenience } \right]  \ If\, \, the\, \, additional\, \, number\, \, of\, \, rotations\, \, is\, \, x,\, \, we\, \, have \ 0={ \left( { { \omega _{ 0\;  } }/2 } \right) _{ \;  } }^{ 2 }\; +\; 2\alpha x\; =\; { \left( { { \omega _{ 0\;  } }/2 } \right) _{ \;  } }^{ 2 }\; +\; 2\times (--\; { \omega _{ 0 } }^{ 2 }/96)x \ This\, \, gives, \ x\; =12 \end{array}$

Hence,
option $(C)$ is correct answer.

The minimum value of ${ \omega  } _{ 0 }$ below which the ring will drop down is 

  1. $\sqrt { \dfrac { g }{ 2\mu (R-r) } } $

  2. $\sqrt { \dfrac { 3g }{ 2\mu (R-r) } } $

  3. $\sqrt { \dfrac { g }{ \mu (R-r) } } $

  4. $\sqrt { \dfrac { 2g }{ \mu (R-r) } } $


Correct Option: C

a flywheel is in the form of a solid circular wheel of mass 72 kg and radius 50cm and it makes 70 r.p.m. then the energy of revolution is:

  1. 245534 J

  2. 24000 J

  3. 4795000J

  4. 4791600 J


Correct Option: D
Explanation:

$K.E=\cfrac{1}{2}mv^2\rightarrow(1)\v=r\omega$

Put in $(1)$
$K.E=\cfrac{1}{2}mr^2\omega^2$
Given data,
$m=72kg\r=50cm\ \omega=70rev/min$
$1rev=2\pi rad\1min=60sec\ \omega=\cfrac{70\times2\pi}{60}=2.33\times3.14\ \omega=7.3rad/sec$
So, $K.E=\cfrac{1}{2}mr^2\omega^2\Rightarrow\cfrac{1}{2}\times72\times50\times50\times\cfrac{7.3}{10}\times\cfrac{7.3}{10}\ K.E=4791600J$

Two discs having masses in the ratio $1:2$ and radii in the ratio $1:8$ roll down without slipping one by one from an inclined plane of height $h$. The ratio of their linear velocities on reaching the ground is

  1. $1:16$

  2. $1:128$

  3. $1:8\sqrt{2}$

  4. $1:1$


Correct Option: D

A disc is rolling on a surface without slipping. What is the ratio of its translational to rotational kinetic energies?

  1. $5:2$

  2. $2:1$

  3. $3:2$

  4. $2:3$


Correct Option: B
Explanation:
Translational Kinetic Energy$=\cfrac{1}{2}mv^{2}=KE _{T}$
Rotational KE$=\cfrac{1}{2}I\omega^{2}$
$v=r\omega\,\,\,\,,I=\cfrac{1}{2}MR^{2}$
Rotational KE$=\cfrac{1}{2}\times\cfrac{1}{2}mR^{2}\times(\cfrac{v}{R})^{2}$
$KE _{r}=\cfrac{1}{4}mv^{2}$
$\cfrac{KE _{T}}{KE _{r}}=\cfrac{\cfrac{1}{2}mv^{2}}{\cfrac{1}{4}mv^{2}}=2:1$

A solid sphere rolls on horizontal surface without slipping. What is the ratio of its rotational to translation kinetic energy.

  1. $\dfrac {2}{5}$

  2. $\dfrac {5}{2}$

  3. $\dfrac {3}{2}$

  4. $0$


Correct Option: D