Tag: stability and centre of mass

Questions Related to stability and centre of mass

Location of centre of mass of uniform semi-circular plate of radius R from its centre is:

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

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

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

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


Correct Option: D

Four bodies of masses 1,2,3,4 kg respectively are placed at the corners of a square of side $'a'$. Coordinates of centre of mass are (take $1\ kg$ at origin, $2\ kg$ on X-axis and $4\ kg$ on Y-axis)

  1. $\Big \lgroup \dfrac{7a}{10}, \dfrac{a}{2} \Big \rgroup$

  2. $\Big \lgroup \dfrac{a}{2}, \dfrac{7a}{10} \Big \rgroup$

  3. $\Big \lgroup \dfrac{a}{2}, \dfrac{3a}{10} \Big \rgroup$

  4. $\Big \lgroup \dfrac{7a}{10}, \dfrac{3a}{2} \Big \rgroup$


Correct Option: C

Two blocks of masses $8$kg are connected by a spring of negligible mass and placed on a frictions less horizontal surface. An impulse gives a velocity of $12$m/s to the heavier block in the direction of lighter block. The velocity of the center of mass is:-

  1. $12$m/s

  2. $10$m/s

  3. $8$m/s

  4. $6$m/s


Correct Option: D
Explanation:

Velocity of center of mass is $=\dfrac{m _1v _1+m _2v _2}{m _1+m _2}$


                                                $=\dfrac{8\times 12+8\times 0}{8+8}$


                                                $=\dfrac{96+0}{16}$

                                                $=6m/s$
Hence, the answer is $6m/s.$

A solid cylinder at rest at the top of an inclined plane of height 2.7 m rolls down without slipping. If the same cylinder has to slide down a frictionless inclined plane and acquire the same velocity as that acquired by the centre of mass of the rolling cylinder at the bottom of the inclined plane, the height of the inclined plane in meters should be 

  1. 2.2

  2. 1.2

  3. 1.6

  4. 1.8


Correct Option: C

Find the coordination of center of mass of a uniform semicircle closed wire frame with respect to the origin which is at its center.The radius of the circular portion is R.                

  1. $\left( {\dfrac{{4R}}{{3\pi }},0} \right)$

  2. $\left( {\dfrac{{2R}}{{\pi }},0} \right)$

  3. $\left( {\dfrac{R}{{\pi + 2}},0} \right)$

  4. $\left( {\dfrac{2R}{{\pi + 2}},0} \right)$


Correct Option: A
Explanation:

We know that, $x+cm=\cfrac{4}{\pi R^2}\int^0 _R{-t^2 dt}$

$=\cfrac{4}{\pi R^2}|-\cfrac{t^3}{R}|^0 _R=\cfrac{4}{\pi R^2}(\cfrac{R^3}{3})=\cfrac{4R}{3\pi}$
Thus, co-ordinates should be $=[\cfrac{4}{3\pi},0]$

Two particles having mass ratio n : 1 are interconnected by a light in extensible string that passes over a smooth pulley. If the system is released, then the acceleration of the centre of mass of the system is

  1. $( n - 1 ) ^ { 2 } g$

  2. $\left( \frac { n + 1 } { n - 1 } \right) ^ { 2 } g$

  3. $\left( \frac { n - 1 } { n + 1 } \right) ^ { 2 } g$

  4. $\left( \frac { n + 1 } { n - 1 } \right) 9$


Correct Option: C
Explanation:

Given$,$

$\frac{{{m _1}}}{{{m _2}}} = \frac{n}{1} = n$
Each mass have the acceleration $a = \frac{{\left( {{m _1} - {m _2}} \right)}}{{{m _1} + {m _2}}}$
however ${{m _1}}$ which is heavier will have the will have acceleration ${{a _1}}$ vertically down while the lighter mass ${{m _2}}$ will have acceleration ${{a _2}}$ vertically up $ \to {a _2} =  - {a _1}$
The acceleration or the centre of mass of the system$,$ ${a _{cm}} = \frac{{{m _1}{a _1} + {m _2}{a _2}}}{{{m _1} + {m _2}}}$
given that ${a _2} =  - {a _1} \to {a _{cm}} = \frac{{\left( {{m _1} - {m _2}} \right){a _1}}}{{{m _1} + {m _2}}} = \frac{{{m _1} - {m _2}}}{{{m _1} + {m _2}}} \times \frac{{\left( {{m _1} - {m _2}} \right)g}}{{{m _1} + {m _2}}} = \frac{{{{\left( {{m _1} - {m _2}} \right)}^2}g}}{{{m _1} + {m _2}}}$
Since $\frac{{{m _1}}}{{{m _2}}} = n$ diving by ${{m _2}}$ and simplifying 
$ \Rightarrow {a _{cm}} = {\left( {\frac{{n - 1}}{{n + 1}}} \right)^2}g$
Hence,
option $(C)$ is correct answer.

A thin uniform rod of length l and m is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is $\omega$. Its centre of mass rises to a maximum height of:

  1. $\dfrac{1}{6} \dfrac{l\omega}{g}$

  2. $\dfrac{1}{2} \dfrac{l^2\omega^2}{g}$

  3. $\dfrac{1}{6} \dfrac{l^2\omega^2}{g}$

  4. $\dfrac{1}{3} \dfrac{l^2\omega^2}{g}$


Correct Option: C
Explanation:

At lowest height

$E=\cfrac{1}{2}I \omega^2$
At maximum weight $\omega=0$
$E=mgh+0$ (his height from lower point)
By conservation of energy
$\cfrac{1}{2}I\omega^2=mgh\h=\cfrac{I\omega^2}{2mg}\h=\cfrac{ml^2\omega^2}{6mg}=\cfrac{l^2\omega^2}{6g}$

Six identical particles each of mass $m$ are arranged at the corners of a regular hexagon of side length $a$. If the mass of one of the particle is doubled, the shift in the centre of mass is

  1. $a$

  2. $\dfrac {6a}{7}$

  3. $\dfrac {a}{7}$

  4. $\dfrac {a}{\sqrt {3}}$


Correct Option: C

Centre of mass is a point 

  1. Which is geometric centre of a body

  2. From which distance of particles are same

  3. Where the whole mass of the body is supposed the

  4. none of these


Correct Option: C

The centre of mass of a rigid body always lies inside the body. Is this statement true or false?

  1. True

  2. False


Correct Option: B
Explanation:
No it's not necessary that the centre of mass of a body should lie inside the body. Consider a circular ring, its centre of mass lies at the center of the ring where there is no content of the body. So it can also lie outside the body .