Tag: centre of gravity

Questions Related to centre of gravity

An object will not undergo rotational motion when:

physics turning effects of forces centre of gravity forces - vectors and moments acceleration due to gravity

  1. the forces are acting on it at different positions

  2. every forces is creating different turning effects

  3. every moment has the same amplitude

  4. all the forces are acting at its centre of gravity

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Correct Option: D
Explanation:

For an object to undergo rotational motion, a net torque must be exerted on the object. If all the forces are acting at its centre of gravity, net torque acting on the object is zero and so the object will not undergo any rotational motion.

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 :

physics turning effects of forces centre of gravity forces - vectors and moments acceleration due to gravity

  1. $(n\,-\,1)^2\, g$

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

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

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

Show answer
Correct Option: C
Explanation:

given $\dfrac { { m } _{ 1 } }{ { m } _{ 2 } }$=$\dfrac{n}{1}$
Each mass will have the acceleration=$a=\dfrac { { (m } _{ 1 }-{ m } _{ 2 })g }{ { 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 -${a} _{2}$=${a} _{1}$
${a} _{c.o.m}$=$\dfrac{{m} _{1}\times{a} _{1}+{m} _{2}\times{a} _{2}}{{m} _{1}+{m} _{2}}$
so ${a} _{c.o.m}$=$\dfrac { { (m } _{ 1 }-{ m } _{ 2 }){a} }{ { m } _{ 1 }+{ m } _{ 2 } } $=$\dfrac{{m} _{1}-{m} _{2}}{{m} _{1}+{m} _{2}}$$\times$$\dfrac { { (m } _{ 1 }-{ m } _{ 2 })g }{ { m } _{ 1 }+{ m } _{ 2 } } $=$\dfrac { { ({ { m } _{ 1 }-{ m } _{ 2 }) }^{ 2 }g } }{ { { (m } _{ 1 }+{ m } _{ 2 }) }^{ 2 } }$
Since $\dfrac{{m} _{1}}{{m} _{2}}$ =n, diving by ${m} _{2}$ and simplifying
${a} _{c.o.m}$=$\dfrac { { (n-1) }^{ 2 } }{ { (n+1) }^{ 2 } }$g

The centre of gravity of the floating ship.

physics turning effects of forces centre of gravity forces - vectors and moments acceleration due to gravity

  1. Coincides with the metacentre

  2. Lies below the metacentre

  3. Lies above the metacentre

  4. None of the above

Show answer
Correct Option: B
Explanation:

The metacentre remains directly above the center of buoyancy regardless of the tilt of a floating body, for example that of the ship. (Center of gravity is the point in a body about which all parts of the body balance each other).

A body weighs 250 N on the surface of the earth. How much will it weigh half way down to the centre of the earth.?

physics turning effects of forces centre of gravity forces - vectors and moments acceleration due to gravity

  1. 125 N

  2. 150 N

  3. 175 N

  4. 250 N.

Show answer
Correct Option: A
Explanation:

Let m be mass of a body.
$\therefore$ Weight of the body on the surface of the earth is W=mg=250N
Acceleration due to gravity at a depth d below the surface of the earth is $g^{\prime}=g\left(1-\dfrac{d}{R _{E}}\right)$
Weight of the body at depth d is
$W^{\prime}=mg^{\prime}=mg\left(1-\dfrac{d}{R _{E}}\right)$
Here, $d=\dfrac{R _{E}}{2}$
$\therefore W^{\prime}=mg\left(1-\dfrac{{R _{E}}/{2}}{R _{E}}\right)=\dfrac{mg}{2}=\dfrac{W}{2}=\dfrac{250N}{2}=125N$

Two planets have radii $r _1$ and $r _2$ and their densities are $\rho _1$ and $\rho _2$, respectively. The ratio of acceleration due to gravity on them will be.

physics turning effects of forces centre of gravity forces - vectors and moments acceleration due to gravity

  1. $r _1\rho _1:r _2\rho _2$

  2. $r _1\rho^2 _1:r _2\rho^2 _2$

  3. $r^2 _1\rho _1:r^2 _2\rho _2$

  4. $r _1\rho _2:r _2\rho _1$

Show answer
Correct Option: A
Explanation:
$\dfrac { { g } _{ 1 } }{ { g } _{ 2 } } =\dfrac { { GM } _{ 1 } }{ { r } _{ 1 }^{ 2 } } \div \dfrac { { GM } _{ 2 } }{ { r } _{ 2 }^{ 2 } } $
$=\dfrac { { Gp } _{ 1 }\dfrac { 4 }{ 3 } \pi { r } _{ 1 }^{ 3 } }{ { r } _{ 1 }^{ 2 } } \div \dfrac { { Gp } _{ 2 }\dfrac { 4 }{ 3 } \pi { r } _{ 2 }^{ 3 } }{ { r } _{ 2 }^{ 2 } } $
$=\dfrac { { p } _{ 1 }{ r } _{ 1 } }{ { p } _{ 2 }{ r } _{ 2 } } $
$=\dfrac { { r } _{ 1 }{ p } _{ 1 } }{ { r } _{ 2 }{ p } _{ 2 } } $
$={ r } _{ 1 }{ p } _{ 1 }:{ r } _{ 2 }{ p } _{ 2 }$

An object after deforming 

physics turning effects of forces centre of gravity forces - vectors and moments acceleration due to gravity

  1. may have different centre of mass.

  2. have same centre  of mass

  3. doesn't change its centre of gravity 

  4. none of the above 

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Correct Option: A
Explanation:

Centre of mass is a point on a physical body which behaves as if the whole mass is concentrated over there. Whenever a force acts on a body it appears to be acting on the centre of mass. 

The centre of mass depends upon how the particles of the body are scattered around it. It is closer towards the heavier regions of the body as compared to the lighter regions.  Whenever an object is deformed, its heavier regions may change their locations. As a result, the centre of mass may change its location

Racing car can corner rapidly without turning due to

physics turning effects of forces centre of gravity forces - vectors and moments acceleration due to gravity

  1. having high centre of gravity

  2. having lower centre of gravity

  3. equal mass distribution of the car

  4. none of the above

Show answer
Correct Option: B
Explanation:

Racing car can corner rapidly without turning due to low centre of gravity. The lower the centre of gravity is more stable is the object The higher it is more likely  the object is to topple over if it is pushed.Racing cars have really low centre of gravity so that they can corner rapidly withoit turning.

Distance covered by a freely falling body in 5 seconds is

physics turning effects of forces centre of gravity forces - vectors and moments acceleration due to gravity

  1. 85 mts

  2. 44 mts

  3. 125 mts

  4. 10 mts

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Correct Option: C

A man can jump 1.5m high on the earth. Calculate the approximate height he might be able to jump on a planet whose density is one-quarter that of the earth and whose radius is one-third of the earth's radius.

physics turning effects of forces centre of gravity forces - vectors and moments acceleration due to gravity

  1. $1.5\ m$

  2. $15\ m$

  3. $18\ m$

  4. $28\ m$

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Correct Option: C

Which among the following is the major cause of nutation in Earth's axis:

physics turning effects of forces centre of gravity forces - vectors and moments acceleration due to gravity

  1. Earth's Oblate spheroid shape

  2. The difference between Land and Sea hemispheres

  3. Oceanic Currents

  4. TIdal Forces

Show answer
Correct Option: D
Explanation:
In case of earth the principal sources of tidal forces are the sun and moon while continuously change location relative to each other and cause nutation in earth's axis, So option (D) correct.