Tag: stability and centre of mass

Questions Related to stability and centre of mass

The reduce mass of two particles having masses m and 2 m is 

  1. 2 m

  2. 3 m

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

  4. $\dfrac { m}{2}$


Correct Option: C
Explanation:

Given,

$m _1=m$
$m _2=2m$

The reduced mass of two particle system is given by
$\mu=\dfrac{m _1m _2}{m _1+m _2}$

$\mu=\dfrac{m\times 2m}{m+2m}$

$\mu=\dfrac{2m}{3}$
The correct option is C.

Two bodies of masses 10 kg and 2 kg are moving with velocities $2\hat { i } -7\hat { k } +3\hat { j }\ m{ s }^{ -1 }$ and $-10\hat { i } +35\hat { k } -3\hat { j }\ m{ s }^{ -1 }$ respectively. The velocity of their centre of mass is

  1. $2\hat { i }\ m{ s }^{ -1 }$

  2. $2\hat { j }\ m{ s }^{ -1 }$

  3. $\left( 2\hat { j } +2\hat { k } \right) m{ s }^{ -1 }$

  4. $\left( 2\hat { i } +2\hat { j } +2\hat { k } \right) m{ s }^{ -1 }$


Correct Option: B
Explanation:
$m _1 = 10\ kg\quad \vec {v _1}=2\hat i -7\hat k+3\hat j\ m/s$
$m _2=2\ kg \quad \vec {v _2}=-10\hat i +35\hat k-3\hat j\ m/s$
velocity of centre of mass should be
$\vec {v}=\dfrac {m _1 \vec {v _1}+m _2 \vec {v _2}}{m _1 +m _2}$
or $\vec {v}=\dfrac {20\hat i-70\hat k+30\hat j+(-20\hat i+70\hat k-6\hat j)}{10+2}$
$\Rightarrow \ \boxed {\vec {v}=\dfrac {24\ \hat j}{12}=2\hat j\ m/s}$

Figure shows a cubical box that has been constructed from uniform metal plat of negligible thickness. The box is open at the top and has edge length $40 /cm$. The $z$ co-ordinate of the centre of mass of the box in $cm$, is  

  1. $12$

  2. $16$

  3. $20$

  4. $22$


Correct Option: B

The centre of mass of a uniform thin hemispherical shell of radius R is located at a distance ?

  1. $\dfrac { \pi R }{ 2 } $

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

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

  4. $\dfrac { 4R }{ 3\pi } $


Correct Option: C
Explanation:

The centre of mass of a uniform thin hemispherical shell of radius R is located at a distance  $\dfrac{R}{2}$ from the center.

A body having its centre of mass at the origin has three of its particles at $(a,0,0),(0,a,0),(0,0,a)$ the moment of inertia of the body about X and Y axis are $0.2kg{m _2}$ the moment of inertia about its Z axis is 

  1. is $0.20kg{m _2}$

  2. is $0.40kg{m _2}$

  3. $0.20\sqrt 2 kg{m^2}$

  4. cannot be deducted with this information


Correct Option: D

The centre of mass of a system of particles is at the origin. It follows that:

  1. the number of particles to the right of the origin is equal to the left of origin.

  2. the total mass of the particles to the right of the origin is same as total mass to the left of the origin.

  3. the number of particles on the X-axis should be equal to the number of particles on the Y-axis .

  4. if there is a particle on the +ve X-axis, there should be atleast one particle on the -ve X-axis.

  5. None of these.


Correct Option: E
Explanation:

Center of mass of a system of particles is at the origin. It follows that:

(a) Number of particles to the right of origin=Number of particles to the left of origin
(b) Total mass of particles to the right of origin=Total mass of particles to the left of origin

The centre of a mass of a rigid body lies

  1. inside the body

  2. outside the body

  3. neither $(a)$ nor $(b)$

  4. either $(a)$ or $(b)$


Correct Option: D
Explanation:

Centre of mass of rigid bodies may lie either inside or outside the body.

For example:  COM of solid sphere lies inside it whereas COM of uniform circular ring lies at its geometric centre where there is actually no matter.

The point through which the total weight appears to act for any orientation of the object is ______.

  1. centre of gravity.

  2. centre of momentum

  3. centre of force

  4. none of the above


Correct Option: A
Explanation:

the point where the total weight appears to act for any orientation of the object is centre of gravity .

The centre of gravity depends on the acceleration due to gravity at the given place.

  1. True

  2. False


Correct Option: B
Explanation:

centre of gravity depends on shape of object and its mass distribution.

so the answer is B.

A uniform metal disc of radius R is taken and out of it a disc of diameter $\dfrac{R}{2}$ is cut off from the end.The centre of mass of the remaining part will be :

  1. $\dfrac{R}{28}$ from the centre

  2. $\dfrac{R}{3}$ from the centre

  3. $\dfrac{R}{5}$ form the centre

  4. $\dfrac{R}{6}$ from the centre


Correct Option: A
Explanation:
No correct option.

Given,

$Radius =R,D=\dfrac{R}{2}$

Let origin be the center  

$0=\dfrac{m _1x _1+m _2x _2}{m _1+m _2}$

Mass of the diameter $\dfrac{R}{2}=\dfrac{m}{8}$

$0=\dfrac{\dfrac{7m}{8}X _1+\dfrac{m}{8}(\dfrac{R}{4})}{\dfrac{7m}{8}+\dfrac{m}{8}}=\dfrac{7m}{8}x _1+\dfrac{m}{8}\times\dfrac{R}{4}\Rightarrow x _1=\dfrac{-R}{28}$