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 A particle of mass m is moving along the side of a square of side ‘a’, with a uniform speed \nu in the x-y plane as shown in the figure : Which of the following statements is false for the angular momentum  \vec{L} about the
origin?
Option: 1 \vec{L}= -\frac{mv}{\sqrt{2}}R\hat{k}when the particle is moving from A to B.

Option: 2 \vec{L}= mv\left [ \frac{R}{\sqrt{2}}-a \right ]\hat{k}when the particle is moving from C to D.

Option: 3 \vec{L}= mv\left [ \frac{R}{\sqrt{2}}+a \right ]\hat{k}when the particle is moving from B to C.

Option: 4 L=\frac{R}{\sqrt{2}} m v(-k)when the particle is moving from D to A.
 

\\ In\ option\ (a), co-ordinates \ of\ A are \left(\frac{R}{\sqrt{2}}, \frac{R}{\sqrt{2}}\right) \\ \therefore \vec{r}=\left(\frac{R}{\sqrt{2}} \hat{i}+\frac{R}{\sqrt{2}} \hat{j}\right) and \vec{v}=v \hat{i}\\ \vec{L} m(\vec{r} \times \vec{v})=m\left(\frac{R}{\sqrt{2}} \hat{i}+\frac{R}{\sqrt{2}} \hat{j}\right) \times v \hat{i}\\ \vec{L}=-\frac{m R}{\sqrt{2}} v \hat{k}

\\ in\ option \ (b)\ it \ moves\ from \ C \ to\ D\\ L=\left(\frac{R}{\sqrt{2}}+a\right) m v(\hat{k})

so option b is correct option

\\ in \ option\ (c), \ For\ B$ \ to\ $C,$ \\ we have $L=\left(\frac{R}{\sqrt{2}}+a\right) m v(\hat{k})

\\ in \ option \ (d), \ When \ a \ particle\ is \ moving\ from \ D \ to\ A\\ L=\frac{R}{\sqrt{2}} m v(-k)

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Posted by

Ritika Jonwal

Consider a uniform rod of massM=4m and length l pivoted about its centre. A mass m moving with velocity v making angle \theta =\frac{\pi }{4} to the rod's long axis collides with one end of the rod and sticks to it. The angular speed of the rod - mass system just after the collision is :   
Option: 1 \frac{4}{7}\frac{v}{l}

Option: 2 \frac{3\sqrt{2}}{7}\frac{v}{l}

Option: 3 \frac{3}{7\sqrt{2}}\frac{v}{l}  

Option: 4 \frac{3}{7}\frac{v}{l}

 

Let us conserve angular momentum about O:-

           So, L_i=\left ( \frac{mv}{\sqrt2} \right )\times \frac{l}{2}, where \left ( \frac{mv}{\sqrt2} \right ) is linear momentum and \left ( \frac{l}{2} \right ) is the distance from centre O.

Now, L_f=I\omega

Here, I=\frac{4ml^2}{12}+{m\left (\frac{l}{2} \right )^2}=\frac{7ml^2}{12}

So, L_i=L_f\Rightarrow \omega=\frac{6v}{7\sqrt2 l}= \frac{3\sqrt2v}{7 l}

So the correct graph is given in option 2.

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Posted by

vishal kumar

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The coordinates of centre of mass of a uniform flag shaped lamina (thin flat plate) of mass 4kg. (The coordinates of the same are shown in the figure) are :
Option: 1 (1.25m,1.50.m)
Option: 2 (0.75m,0.75m)
Option: 3 (0.75m,1.75m)
Option: 4 (1m,1.75m)
 

 

 

 

 

The co-ordinate of O1 is (0.5,1),  O2 is (1, 2.5)

\\x_{cm}=\frac{m\times0.5+m\times1}{2m}=0.75\\ y_{cm}=\frac{m\times+m\times\frac{5}{2}}{2m}=1.75

So the correct option is 3.

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Posted by

vishal kumar

Three point particles of masses 1.0 kg, 1.5 kg and 2.5 kg are placed at three corners of a right triangle of sides 4.0 cm, 3.0 cm, and 5.0 cm as shown in the figure. The centre of mass of the system is at a point:
 

Option: 1 2.0\; cm right and 0.9\; cm above 1\: kg mass
   
Option: 2 0.9\; cm right and 2.0\; cm above 1\: kg mass

Option: 3 0.6\; cm right and 2.0\; cm above 1\: kg mass  

Option: 4 1.5\; cm right and 1.2\; cm above 1\: kg mass
 

 

 

let m1=1 kg, m2=1.5 kg and m3=2.5 kg

x1=0, x2=3, x3=0 and y1=0, y2=0, y3=4

x_{com}=\frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3} and y_{com}=\frac{m_1y_1+m_2y_2+m_3y_3}{m_1+m_2+m_3}

Let point A be origin and mass m1=1.0 kg be at origin.

So, x_{com}=\frac{1\times0+1.5\times3+2.5\times0}{1+1.5+2.5}=\frac{4.5}{5}=0.9

and y_{com}=\frac{1\times0+1.5\times0+2.5\times4}{1+1.5+2.5}=\frac{10}{5}=2

so centre of mass of the system is at (0.9,2).

So from the figure we can say that the 0.9 cm right and 2 cm above the 1 kg mass. 

So option (2) is correct.

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Posted by

Ritika Jonwal

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As shown in the figure, a bob of mass m is tied by a massless string whose other end portion is wound on a flywheel (disc) of radius r and mass m. When released from rest the bob starts falling vertically. When it has covered a distance h, the angular speed of the wheel will be:-
 
Option: 1 r\sqrt{\frac{3}{2gh}}
 
Option: 2 r\sqrt{\frac{3}{4gh}}

Option: 3 \frac{1}{r}\sqrt{\frac{4gh}{3}}  

Option: 4 \frac{1}{r}\sqrt{\frac{2gh}{3}}
 

 

 

 

Conservation Of angular momentum -    

\\mg-T=ma\\T\times r=I\alpha\\T=\frac{mr^2}{2}\times\frac{a}{r}\times\frac{1}{r}\\T=\frac{ma}{2}\\mg=\frac{3ma}{2}\\a=\frac{2g}{3}\\Also,\ v=\sqrt{2as}=\sqrt{\frac{4gh}{3}}\\also, v=\omega r\\ \omega=\frac{v}{r}

\Rightarrow \sqrt{\frac{4gh}{3}}\times\frac{1}{r}=\frac{1}{r}\sqrt{\frac{4gh}{3}}

Hence option (3) is correct.

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Posted by

Ritika Jonwal

The radius of gyration of a uniform rod of length l, about an axis passing through a point \frac{l}{4} away from the centre of the rod and perpendicular to it is:-  
Option: 1 \frac{1}{8}l

 
Option: 2 \sqrt{\frac{7}{48}l}

Option: 3 \sqrt{\frac{3}{8}l}  

Option: 4 \frac{1}{4}l
 

 

 

                  

Moment of inertia of rod about an axis perpendicular through COM

I_{COM}=\frac{Ml^2}{12}

And I_N=I_{COM}+M(\frac{l}{4})^2=\frac{Ml^2}{12}+M(\frac{l^2}{16})=\frac{7Ml^2}{48}

Radius of Gyration

\\ K=\sqrt{\frac{I_N}{M}}=\sqrt{\frac{7Ml^2}{48M}}=\frac{l}{4}*\sqrt{\frac{7}{3}} \\ = l\sqrt{\frac{7}{48}}

 

So the option (2) is correct.

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Posted by

Ritika Jonwal

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A rod 'l' has non-uniform linear mass density given by \rho (x)=a+b\left ( \frac{x}{l} \right )^{2}, where a and b are constants and 0\leq x\leq l. The value of x for the centre of mass of the rod is at :
Option: 1 \frac{3}{2}\left ( \frac{2a+b}{3a+b} \right )l
Option: 2  \frac{3}{2}\left ( \frac{a+b}{2a+b} \right )l
Option: 3  \frac{3}{4}\left ( \frac{2a+b}{3a+b} \right )l
Option: 4 \frac{4}{3}\left ( \frac{a+b}{2a+3b} \right )l
 

\begin{aligned} &\text { Given, } \lambda=\left(\mathrm{a}+\mathrm{b}\left(\frac{\mathrm{x}}{\mathrm{l}}\right)^{2}\right) \\ &\frac{\mathrm{d} \mathrm{N}}{\mathrm{dx}}=\lambda \\ &\mathrm{d} \mathrm{N}=\lambda \mathrm{dx}=\left(\mathrm{a}+\mathrm{b}\left(\frac{\mathrm{x}}{\mathrm{l}}\right)^{2}\right) \cdot \mathrm{dx} \\ &=\frac{\int \mathrm{x} \cdot \mathrm{dm}}{\int \mathrm{dm}}=\frac{\int \mathrm{x} \lambda \mathrm{dx}}{\int \lambda \mathrm{dx}} \\ &=\frac{\int_{0}^{1} \mathrm{x}\left(\mathrm{a}+\left(\frac{\mathrm{bx}^{2}}{1^{2}}\right)\right) \mathrm{dx}}{\int_{0}^{1}\left(\mathrm{a}+\frac{\mathrm{bx}^{2}}{1^{2}}\right) \mathrm{dx}}=\frac{\int_{0}^{1}\left(\mathrm{ax}+\frac{\mathrm{bx}^{3}}{1^{2}}\right) \mathrm{dx}}{\int_{0}^{1}\left(\mathrm{a}+\frac{\mathrm{bx}^{2}}{1^{2}}\right) \mathrm{dx}} \end{aligned}

\begin{aligned} &=\frac{\left[\frac{a x^{2}}{2}\right]_{0}^{1}+\frac{b}{1^{2}}\left[\frac{x^{4}}{4}\right]_{0}^{1}}{a[x]_{0}^{1}+\frac{b}{1^{2}}\left[\frac{x^{3}}{3}\right]_{0}^{1}}=\frac{\frac{a^{1^{2}}}{2}+\frac{b^{1^{2}}}{4}}{a l+\frac{b l}{3}} \\ &=\frac{(2 a+b) 1}{(3 a+b) 4} \times 3 \\ &=\frac{31}{4}\left(\frac{2 a+b}{3 a+b}\right) \end{aligned}

 

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Posted by

avinash.dongre

A body mass m = 10 kg is attached to one end of a wire of length 0.3m. The maximum angular speed ( in rad s-1) with which it can be rotated about its other end in space station is (Breaking stress of wire =4.8\times 10^7 Nm^{-2} and area of cross-section of the wire =10^{-2} cm^2) is:
Option: 1 4 
Option: 2 8
Option: 3 6
Option: 4 1
 

 

\begin{aligned} \frac{F}{A} &=\frac{m v^{2}}{l A} \\ \Rightarrow v &=\sqrt{\frac{l F}{m} }=\sqrt{\frac{0.3}{10} \times 4.8 \times 10^{7} \times 10^{-6}} \\ &=\sqrt{3 \times 48 \times 10^{4} \times 10^{-6}}=1.2 \mathrm{m} / \mathrm{s} \\ \omega &=\frac{v}{l}=\frac{1.2}{0.3}=4 \mathrm{rad} / \mathrm{s} \end{aligned}

 

the correct option is (1). 

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Posted by

avinash.dongre

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One end of a straight uniform 1m long bar is pivoted on a horizontal table. It is released from rest when it makes an angle 30o from the horizontal (see figure). Its angular speed when it hits the table is given as \sqrt n s^{-1}, where n is an integer. The value of n is __________.
Option: 1 15
Option: 2 12
Option: 3 24
Option: 4  30

mg\frac{t}{2}\sin 30\degree = \frac{1}{2}\frac{mt^2}{3}\omega^2

Solving

\omega^2 = 15 \\\Rightarrow \omega = \sqrt{15}

Since, \omega=\sqrt{n}= \sqrt{15} , Therefore n= 15.

Hence the correct option is (1). 

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Posted by

avinash.dongre

A uniform sphere of mass 500 g rolls without slipping on a plane horizontal surface with its center moving at a speed of 5.00 cm/s. Its kinetic energy is :
Option: 1 6.25\times 10^{-4} J
Option: 2 1.13\times 10^{-3} J
Option: 3 8.75\times 10^{-4} J
Option: 4 8.75\times 10^{-3} J
 

 

 

 

 

 

 

The kinetic energy of sphere is kinetic energy of translation+ kinetic energy of rotation

K=\frac{1}{2}mv^2(1+\frac{k^2}{R^2})

Where K is the radius of gyration

\\K=\frac{1}{2}\times0.5(\frac{5}{100})^2(1+\frac{2}{5})\\K=\frac{35}{4}\times10^{-4}J=8.75 \times10^{-4}J

Hence the correct option is (3).

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Posted by

vishal kumar

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