Physics

Gravitation and Center of Mass

368 Questions

Gravitation and center of mass questions explore gravitational fields, planetary density, and the mechanics of celestial bodies. Test items include calculating gravitational strength on different planets and understanding the Roche Limit. This topic is essential for the physics syllabus of major competitive exams.

Gravitational fieldCenter of massPlanetary densityHill SphereSpace-time curvature

Gravitation and Center of Mass Questions

Multiple choice physics turning effects of forces stability and centre of mass center of mass centre of mass

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

Multiple choice physics turning effects of forces stability and centre of mass center of mass centre of mass

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

Multiple choice physics turning effects of forces stability and centre of mass center of mass centre of mass

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $\dfrac{R}{2}=\dfrac{m}{8}$

$0=\dfrac{\dfrac{7m}{8}X _1+\dfrac{m}{8}(\dfrac{R}{4})}{\dfrac{7m}{8}+\dfrac{m}{8}}=\dfrac{7m}{8}x _1+\dfrac{m}{8}\times\dfrac{R}{4}\Rightarrow x _1=\dfrac{-R}{28}$
Multiple choice physics turning effects of forces stability and centre of mass center of mass 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. $\dfrac{4R}{3\pi}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The center of mass of a uniform semi-circular plate of radius R is located on the axis of symmetry at a distance of 4R / (3 * pi) from the center of the base.

Multiple choice physics turning effects of forces stability and centre of mass center of mass centre of mass

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. $\Big \lgroup \dfrac{7a}{10}, \dfrac{3a}{2} \Big \rgroup$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics turning effects of forces stability and centre of mass center of mass centre of mass

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. $\dfrac {a}{\sqrt {3}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The center of mass of a regular hexagon with identical masses is at the center. If one mass m is doubled to 2m, the system is equivalent to the original system plus an additional mass m at that corner. The shift is (m * r) / (total mass), where r is the distance from the center. Total mass = 7m. Shift = (m * a) / 7m = a/7.

Multiple choice physics turning effects of forces stability and centre of mass center of mass centre of mass

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
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The center of mass is the unique point where the weighted relative position of the distributed mass sums to zero. It is the point at which the entire mass of the body can be considered to be concentrated for the purpose of translational motion analysis.

Multiple choice physics turning effects of forces stability and centre of mass center of mass centre of mass

A body having it's center of mass  at the origin. Then,

(The question having a multiple answers).

  1. x co-ordinates of the particles may be all positive.

  2. total KE must be conserved.

  3. total KE must very.

  4. total momentum shall vary.
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation

If all the particles have positive x  co-ordinates then their COM can't be at the origin. Total KE may not be conserved because of internal forces. Its not given that there is no external force acting on the system, so its momentum may also change.

Multiple choice physics turning effects of forces stability and centre of mass center of mass centre of mass

Where will be the centre of mass on combining two masses $m$ and $M(M>m)$ ?

  1. $Towards \ m$
  2. $Towards \ M$
  3. $ exactly \ between \ m \ and \ M $
  4. $None \ of \ the \ above$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

As we can see that the center of mass will be at $r _c=\dfrac{mr+MR}{m+M}$,


where $r$ and $R$ are the position of the masses $m$ and $M$ respectively.

From the formula we notice that it will be between $m$ and $M$ and $nearer $ to the higher mass $M$.

Multiple choice physics turning effects of forces stability and centre of mass center of mass centre of mass

A body has its center of mass at the origin. The x-axis coordinates of the particles :

  1. may be all positive

  2. may be all negative

  3. should be all at zero

  4. may be positive for some case and negative in other cases.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

All co-ordinates positive will make the co-ordinate of com positive.
All co-ordinates negative will make the co-ordinates of com negative.
If all the co-ordinates are zero(0), co-ordinate of com is zero(0).
If  co-ordinate of com can be zero,  co-ordinates of some are positive and co-ordinates of some are negative.

Multiple choice physics turning effects of forces stability and centre of mass center of mass centre of mass

If the linear density of a rod of length L varies as $\lambda =A+B _x$, compute its centre of mass.

  1. $[\cfrac {L(3A+2BL)}{3(2A+BL},0,0]$
  2. $[0,\cfrac {(3A+2B)L}{(2A+3L},\cfrac L 2]$
  3. $[0,0\cfrac {L(3A+2BL)}{3(2A+BL}]$
  4. $[\cfrac L 2,00]$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The center of mass for a non-uniform rod is found by integrating x*lambda*dx divided by the total mass. With lambda = A + Bx, the integral of x(A+Bx) from 0 to L yields the numerator, and the integral of (A+Bx) yields the denominator.

Multiple choice physics turning effects of forces stability and centre of mass center of mass centre of mass

If a square of side $\dfrac{R}{2}$ is removed from a uniform circular disc of radius R as shown in the figure, the shift in centre of mass is 

  1. $\dfrac{R}{4 \pi - 1}$
  2. $\dfrac{R}{2(4 \pi - 1)}$
  3. $\dfrac{R}{3(4 \pi - 1)}$
  4. $\dfrac{R}{4(4 \pi - 1)}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The shift in the center of mass is calculated using the principle of negative mass. The mass of the removed square is proportional to its area, and the shift is (m_removed * distance) / (M_total - m_removed).

Multiple choice physics turning effects of forces stability and centre of mass center of mass centre of mass

Which of the following is not correct about centre of mass ?

  1. It depends on the choice of frame of reference

  2. In centre of mass frame, momentum of a system is always zero

  3. Internal forces may affect the motion of centre of mass

  4. Centre of mass and centre of gravity coincide in uniform gravitational field
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The internal forces all balance each other so that there is no change in the mass distribution of the body, thus center of mass remains unchanged. Rest all options are correct.