Tag: the turning effect of a force

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Questions Related to the turning effect of a force

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

A small piece of space junk is at rest in outer space. A very small asteroid strikes it, exerting a force on it that is NOT directed through the piece of space junk's center of mass.
Which of the following describes the motion of the piece of space junk DURING the asteroid strike?

  1. Because the asteroid is small, the space junk remains at rest

  2. The piece of space junk spins, but does NOT move linearly

  3. The piece of space junk moves at constant velocity linearly, but does NOT spin

  4. The piece of space junk accelerates linearly, but does NOT spin

  5. The piece of space junk accelerates linearly, AND spins
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Due to striking the space junk, the asteroid will exert a force on the junk piece.

This force is not directed towards the of mass of the junk.
Hence the junk would experience a linear acceleration given by $a=\dfrac{F}{m}$
And also the angular acceleration as provided by the force=$\dfrac{Fl}{I}$
where $l$ is the least distance between the center of mass of junk and line along which asteroid moves,
and $I $ is the moment of inertia of the space junk.

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

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?

  1. $1m$

  2. $0.4m$

  3. $0.25m$

  4. zero
Reveal answer Fill a bubble to check yourself
D Correct answer
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.   

Multiple choice 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

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

  1. inertia

  2. equilibrium

  3. suspension

  4. motion
Reveal answer Fill a bubble to check yourself
B Correct answer
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

Multiple choice 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

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

  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. $A-p,q,r;B-q;r;C-s;p;D-p,q,r,s$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice 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

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.
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice 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

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. $18$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

Multiple choice 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

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) } } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

This relates to the critical angular velocity required for a ring to maintain contact or prevent slipping in a rotating system. The derivation leads to the expression in option C.

Multiple choice 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

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
Reveal answer Fill a bubble to check yourself
D Correct answer
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, $K.E=\cfrac{1}{2}mr^2\omega^2\Rightarrow\cfrac{1}{2}\times72\times50\times50\times\cfrac{7.3}{10}\times\cfrac{7.3}{10}\ K.E=4791600J$

Multiple choice 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

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. $1:1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a disc rolling down an incline, the final linear velocity v = sqrt(2gh / (1 + k^2/R^2)). For a uniform disc, k^2/R^2 = 0.5. Since the velocity depends only on height h and the shape (moment of inertia factor), the mass and radius do not affect the final velocity. Thus, the ratio is 1:1.

Multiple choice physics turning effects of forces the turning effect of a force moment of force or torque couple

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. $2:3$
Reveal answer Fill a bubble to check yourself
B Correct answer
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$=\cfrac{1}{2}\times\cfrac{1}{2}mR^{2}\times(\cfrac{v}{R})^{2}$
$KE _{r}=\cfrac{1}{4}mv^{2}$
$\cfrac{KE _{T}}{KE _{r}}=\cfrac{\cfrac{1}{2}mv^{2}}{\cfrac{1}{4}mv^{2}}=2:1$