Tag: turning on a pivot

Questions Related to turning on a pivot

A body is under the action of two equal and oppositely directed forces and the body is rotating with constant non-zero angular acceleration. Which of the following cannot be the separation between the lines of action of the forces?

physics turning on a pivot the turning effect of a force moment of force or torque turning effect of force couple

  1. $1m$

  2. $0.4m$

  3. $0.25m$

  4. zero

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Correct Option: D
Explanation:

A pair of equal and opposite forces acting on a body is called a couple.

It only gives a rotating effect to the body and the torque due to a couple is
given by $\tau = Fd$
where $F$ is equal to the  magnitude of one of the forces and 
$d$ is the distance between their points of application.
To produce a non-zero angular acceleration , the torque on the body must be non zero.
Hence , the couple can give a non-zero angular acceleration to the body
only when the distance between the points of applications of the forces is non-zero.   

 If principle of moments for any object holds, then object is in state of

physics turning on a pivot the turning of couple couple turning effect of force the turning effect of a force moment of force or torque

  1. inertia

  2. equilibrium

  3. suspension

  4. motion

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Correct Option: B
Explanation:

If principle of moments hold good, then the net torque about a given point is zero (usually CM or the pivoted point is zero). Hence the object does not rotate and is said to be in equilibrium

A uniform dice of mass $10kg$ radius $1m$ is placed on a rought horizontal surface. The coefficient of friction between the disc and the surface is $0.2$. A horizontal time varying force is applied on the centre of the disc whose variation with time is shown in graph.
List-I                                                         List-IIDisc rolls without slipping                   at $t=7s$Disc rolls with slipping                       at $t=3s$  Disc starts slipping at                         at $t=4s$Friction force is $10N$ at              None

physics turning on a pivot the turning of couple couple turning effect of force the turning effect of a force moment of force or torque

  1. $A-p,q;B-p;C-r;Dq$

  2. $A-p,r;B-s;C-s,p;D-q$

  3. $A-q,r;B-p;C-s;D-q$

  4. $A-p,q,r;B-q;r;C-s;p;D-p,q,r,s$

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Correct Option: C

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 ?

physics turning on a pivot the turning of couple couple turning effect of force the turning effect of a force moment of force or torque

  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.

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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)

physics turning on a pivot the turning of couple couple turning effect of force the turning effect of a force moment of force or torque

  1. $48$

  2. $36$

  3. $12$

  4. $18$

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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 

physics turning on a pivot the turning of couple couple turning effect of force the turning effect of a force moment of force or torque

  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) } } $

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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:

physics turning on a pivot the turning of couple couple turning effect of force the turning effect of a force moment of force or torque

  1. 245534 J

  2. 24000 J

  3. 4795000J

  4. 4791600 J

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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

physics turning on a pivot the turning of couple couple turning effect of force the turning effect of a force moment of force or torque

  1. $1:16$

  2. $1:128$

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

  4. $1:1$

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Correct Option: D