Tag: turning effects of forces

Questions Related to turning effects of forces

If linear density of a rod of length 3 m varies as  $\lambda=2+x$, then the position of the centre of gravity of the rod is 

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

  1. $\dfrac{7}{3}m$

  2. $\dfrac{12}{7}m$

  3. $\dfrac{10}{7}m$

  4. $\dfrac{9}{7}m$

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Correct Option: B

If R is radius of the planet and g is the acceleration due to gravity at its surface then the body will reach the centre of the planet in time 

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

  1. 2$\pi$$\sqrt\frac{R}{g}$

  2. $\pi$$\sqrt\frac{R}{g}$

  3. $\frac{\pi}{2}$$\sqrt\frac{R}{g}$

  4. $\sqrt\frac{2R}{g}$

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Correct Option: B

A bullet of mass 0.01$\mathrm { kg }$ and traveling at a speed of 500$\mathrm { m } / \mathrm { sec }$ strikes a block of which suspended by a string of length 5$\mathrm { m }$ . The centre of gravity of the block is found to vertical distance of 0.1$\mathrm { m }$ . What is the speed of the bullet after it emerges from the block?

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

  1. 359$\mathrm { m } / \mathrm { s }$

  2. 220$\mathrm { m } / \mathrm { s }$

  3. 204$\mathrm { m } / \mathrm { s }$

  4. 284$\mathrm { m } / \mathrm { s }$

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Correct Option: A
Explanation:

By energy conservation of the suspended block we can say 

initial $KE=$ final potential energy
$\dfrac{1}{2}m{v^2} = mgH$
$\dfrac{1}{2}m{v^2} = gH$
$v = \sqrt {2gH} $
$v = \sqrt {2 \times 10 \times 0.1}  = 1.41\,m/s$
now$,$ by momentum conservation 
${P _{bullet}} = {P _{block}} + {P _{bullet}}$
$0.01 \times 500 = 1 \times 1.41 + 0.01 \times {v _f}$
$5 - 1.41 = 0.01 \times {v _f}$
$0.01{v _f} = 3.59$
${v _f} = 359\,m/s$
Hence,
option $(A)$ is correct answer.

A metallic rod falls under gravity with its ends pointing east and west. Then

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

  1. an emf is induced in it as it cuts the magnetic lines of force

  2. no emf is induced at all

  3. two emf of equal nut opposite signs are induced, giving no emf is

  4. its acceleration is equal to the product of g and the radius of the rod.

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Correct Option: A
Explanation:

When rod falls due to earth magnetic field an emf is induced 

Hence,
option $A$ is correct answer.

The apparent weight of a person of mass m in an elevator is 2mg. The elevator is moving

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

  1. up with an acceleration of $\frac{g}{2}$

  2. up with an acceleration of g

  3. up with an acceleration of 2g

  4. down with an acceleration of g

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Correct Option: B

A body of mass 5 kg initially moving with speed 10 m/s along x-axis in gravity free space explodes and breaks into three pieces of masses 1 kg, 1 kg and 3 kg. the two pieces of equal masses fly off with the same speed 20 m/s along y-axis and z-axis respectively. what is the velocity of heavier fragment?

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

  1. $ ( \frac {10}{3} \hat i - \frac {20}{3} \hat j - \frac {40}{3} \hat k ) m/s $

  2. $ ( \frac {50}{3} \hat i - \frac {20}{3} \hat j - \frac {20}{3} \hat k ) m/s $

  3. $ ( \frac {20}{3} \hat i - \frac {20}{3} \hat j - \frac {20}{3} \hat k ) m/s $

  4. None

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Correct Option: A

AU the particles of a system are situated at a distance r from the origin. The distance of the centre of mass of the system from the origin is :

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

  1. = r

  2. $\leq \, r$

  3. $> r$

  4. $\leq 0$

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Correct Option: B
Explanation:

The largest distance between the origin and center of mass would be 'r' when there is only one particle.
If there are more than one particle, the center of mass would be inside the circle of radius 'r' centered at the origin.
Hence option B is correct.

The position vector of three particles of masses $m _1\, =\,1kg,\, m _2\, =\, 2\, kg$ and $m _3\, =\, 3\, kg$ are $\vec{r} _1\, =\, (\hat{i}\, +\, 4\hat{j}\, +\, \hat{k})\, m,\, \vec{r} _2\, =\, (\hat{i}\, +\, \hat{j}\, +\, \hat{k}) m$ and $\vec{r} _3\, =\, (2\hat{i}\, -\, \hat{j}\, -\, 2\hat{k})$ m respectively. Find the position vector of their center of mass.

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

  1. $\displaystyle \frac {1}{2}\, (\hat{i}\, +\, \hat{j}\, -\, \hat{k})\, m$

  2. $\displaystyle \frac {1}{2}\, (\hat{i}\, +\, 3\hat{j}\, -\, \hat{k})\, m$

  3. $\displaystyle \frac {1}{2}\, (\hat{i}\, +\, \hat{j}\, -\, 3\hat{k})\, m$

  4. $\displaystyle \frac {1}{2}\, (3\hat{i}\, +\, \hat{j}\, -\, \hat{k})\, m$

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Correct Option: D
Explanation:

The position vector of COM of the.three particles will be given by
$\vec{r} _{COM}\, =\, \displaystyle \frac {m _1\vec{r} _1\, +\, m _2\vec{r} _2\, +\, m _3\vec{r} _3}{m _1\, +\, m _2\, +\, m _3}$
Substituting the values, we get
$\vec{r} _{COM}\, =\, \displaystyle \frac {(1) (\hat{i}\, +\, 4\hat{j}\, +\, \hat{k})\, +\, (2) (\hat{i}\, +\, \hat{j}\, +\, \hat{k})\, +\, (3) (2\hat{i}\, -\, \hat{j}\, -\, 2\hat{k})}{1+2+3}\, =\, \displaystyle \frac {1}{2}\, (3\hat{i}\, +\, \hat{j}\, -\, \hat{k})\, m$.
Hence, the position vector of their center of mass is $\, \displaystyle \frac {1}{2}\, (3\hat{i}\, +\, \hat{j}\, -\, \hat{k})\, m$.

At which point is the centre of gravity situated in: A circular lamina.

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

  1. At the centre of radius.

  2. At the centre of semi circular lamina.

  3. At the centre of circular lamina.

  4. can not say

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Correct Option: C
Explanation:

Centre of gravity means a point from which the weight of a body or system may be considered to act. In uniform gravity it is the same as the centre of mass. For regular bodies centre of gravity lies at the centre of the body. Hence this will be at the centre of the circular lamina.

What is the position of centre of gravity of a rectangular lamina?

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

  1. At the mid point of longer side

  2. At the mid point of shorter side

  3. At the point of intersection of its diagonals

  4. At one of the corners

Show answer
Correct Option: C
Explanation:

Centre of gravity means a point from which the weight of a body or system may be considered to act. In uniform gravity it is the same as the centre of mass. Hence for a regular shaped bodies it will have at the centre of that body. Hence for a rectangle it is nothing but at the point of intersection of diagonals.