Tag: turning effects of forces

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Questions Related to turning effects of forces

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

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
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $6m/s.$

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

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
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

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. $\left( {\dfrac{2R}{{\pi + 2}},0} \right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $=[\cfrac{4}{3\pi},0]$

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

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. $\left( \frac { n + 1 } { n - 1 } \right) 9$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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. $\dfrac{1}{3} \dfrac{l^2\omega^2}{g}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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
$\cfrac{1}{2}I\omega^2=mgh\h=\cfrac{I\omega^2}{2mg}\h=\cfrac{ml^2\omega^2}{6mg}=\cfrac{l^2\omega^2}{6g}$

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.