Tag: rotational motion of a rigid body and moment of inertia

Questions Related to rotational motion of a rigid body and moment of inertia

A particle starts from the point 0.8 m and moves with a uniform velocity of 3 m/s. what is the angular momentum of the particle after 5 seconds about origin.mass of the particle is 1  kg.

  1. $12kgm^2$/s

  2. $24 kgm^2/s$

  3. $32 kgm^2/s$

  4. $44 kgm^2/s$


Correct Option: B
Explanation:

The correct option is B

Given,

$mass=1kg$

$t=5s$

$velocity=3m/s$

The angular momentum of the particle about a fixed position is given by:

$L = r \times mV$

So it is the product of momentum of a particle and its perpendicular distance from the fixed position.


Here particle is moving along the x-direction



So its momentum is along the x-direction

$P = mv = 1\times3 = 3 kgm/s$

Now, the perpendicular distance from origin to the momentum of particle is:

$r = 8 m$

Now, angular momentum is


$L = 8\times3 = 24 kg m^2/s$

So angular momentum is $ 24 kg m^2/s$

The instantaneous angular position of a point on a rotating wheel is given by the equation $\theta(t)=2t^{3}-6t^{2}$. The torque on the wheel becomes zero at :

  1. $t=1s$

  2. $t=0.5\ s$

  3. $t=0.25\ s$

  4. $t=2s$


Correct Option: A
Explanation:
Given, $\theta (t)=2t^3-6t^2$

We, know $T=\propto \dfrac{Id^2 \theta}{dt^2}=I\dfrac{d}{dt}(\dfrac{d \theta}{dt})=I\dfrac{d}{dt}(6t^2-12t)=I(12t-12)=0$

Then, $12t-12=0\Rightarrow t=1s$

Let $A$ be the area swept by the line joining the earth and the sun during Feb 2012. The area swept by the same line during the first week of that month is

  1. $\dfrac{A}{4}$

  2. $\dfrac{7A}{29}$

  3. $A$

  4. $\dfrac{7A}{30}$


Correct Option: B

(i) Linear momentum is proportional to $1/n$
(ii) Radius is proportional to $n$
(iii) Kinetic energy is proportional to $1/n^{2}$
(iv) Angular momentum is proportional to $n$
Choose the correct option from the codes given below. 

  1. $(i),(iii),(iv)$ are correct

  2. $(i)$ is correct

  3. $(i),(ii)$ are correct

  4. $(iii)$ is correct


Correct Option: C

The position of a particle is given by $\overrightarrow { r } =(\hat{i} +2\hat{j} -\hat{k})$ and momentum $\overrightarrow { P } =(3\hat{i} +4\hat{j} -2\hat{k})$. The angular momentum is perpendicular to

  1. X-axis

  2. Y-axis

  3. Z-axis

  4. Line at equal angles to all the three axes


Correct Option: A
Explanation:

Given,

$\vec r=\hat i+2\hat j-\hat k$
$\vec P=3\hat i+4\hat j-2\hat k$
Angular momentum, $\vec L=\vec r \times \vec P$
$\vec L=(\hat i+2\hat j-\hat k)\times (3\hat i+4\hat j-2\hat k)$
$\vec L=0\hat i-\hat j-2\hat k$
Hence, The angular momentum is perpendicular to the X-axis.
The correct option is A.

A particle of mass $1\ kg$ is projected at an angle $ \theta $ with horizontal. Its co-ordinates at any instant are $(5m,5m)$ and itis having velocity components along $X- $axis and $Y-$axis as $8\ m/s$ and $4\ m/s$ respectively. Its angular momentum about the origin is 

  1. $- 20\ N-m$ $ \hat{k} $

  2. $+ 20\ N-m$ $ \hat{k} $

  3. $- 60\ N-m$ $ \hat{k} $

  4. $+ 60\ N-m$ $ \hat{k} $


Correct Option: B

A particle of mass m = 5 units is moving with a uniform speed v = $3\sqrt2$ units in the XY - plane along the line y = x + 4.The magnitude of the angular momentum about origin is  

  1. zero

  2. 60 unit

  3. 7.5 unit

  4. $40\sqrt2$ unit


Correct Option: B

The angular momentum of an electron in a hydrogen atom is proprotional to ( where r is redius of orbit) 

  1. $\frac{1}{\sqrt{r}}$

  2. $\frac{1}{r}$

  3. $r^{1/2}$

  4. $^r{2}$


Correct Option: C
Explanation:

Radius of the nth orbit rn n2/Z          n (r n)½

Angular momentum Ln = nh/(2π) (r n)½

$=r^{1/2}$

If a particle of mass m is moving with constant velocity V parallel to X-axis along $y=axis$. The angular momentum with respect to origin at any time 't' is

  1. $mV\hat{J}$

  2. $-mVa\hat{K}$

  3. $mV\hat{K}$

  4. $mVa\hat{J}$


Correct Option: A

The rotational kinetic energy of a hollow spherical shell 2.5 J. If its frequency of rotation is made 10 times, then new kinetic energy will be -

  1. $0.25 J$

  2. $2.5\times { 10 }^{ 2 }J$

  3. $2500 J$

  4. $2.5 J$


Correct Option: B
Explanation:
The rotational kinetic energy is given by
$K=\dfrac{1}{2}mr^2\omega ^2=2.5 J$. . . . .(1)
If frequency of rotation made 10 times , then the new rotational kinetic energy is
$K'=\dfrac{1}{2}mr^2 \times 10^2\omega ^2$
$K'=100K$ 
$K'=2.5\times 10^2J$
The correct option is B.