5 Marks Questions

Question 15 Marks

Two small balls A and B, each of mass m, are joined rigidly to the ends of a light rod of lengh L (figure). The system translates on a frictionless horizontal surface with a velocity vo in a direction perpendicular to the rod. A particle P of mass m kept at rest on the surface sticks to the ball A as the ball collides with it. Find:

The linear speeds of the balls A and B after the collision.
The velocity of the centre of mass C of the system A + B + P.
The angular speed of the system about C after the collision.

[Hint: The light rod will exert a force on the ball B only along its length.]

Answer

Two balls A & B, each of mass m are joined rigidly to the ends of a light of rod of length L. The system moves in a velocity v₀ in a direction $\perp$ to the rod. A particle P of mass m kept at rest on the surface sticks to the ball A as the ball collides with it.

The light rod will exert a force on the ball B only along its length. So collision will not affect its velocity.

B has a velocity = v₀

If we consider the three bodies to be a system.

Applying L.C.L.M.

Therefore $\text{mv}_0=2\text{mv}'$

$\Rightarrow\text{v}'=\frac{\text{v}_0}{2}$

Therefore a has velocity $\frac{\text{v}_0}{2}$

if we consider the three bodies to be a system.

Therefore, net external force = 0

Therefore $\text{V}_{\text{cm}}=\frac{\text{m}\times\text{v}_0+2\text{m}\big(\frac{\text{v}_0}{2}\big)}{\text{m}+\text{2m}}$

$=\frac{\text{mv}_0+\text{mv}_0}{3\text{m}}=\frac{2\text{v}_0}{3}$ (along the initial velocity as before collision)

The velocity of (A + P) w.r.t. the centre of mass $=\frac{2\text{v}_0}{3}-\frac{\text{v}_0}{2}=\frac{\text{v}_0}{6}$ &

The velocity of B w.r.t. the centre of mass $\text{v}_0-\frac{2\text{v}_0}{3}=\frac{\text{v}_0}{3}$

[Only magnitude has been taken]

Distance of the (A + P) from centre of mass $=\frac{\text{l}}3{}$ & for B it is $=\frac{2\text{l}}3{}.$

Therefore $\text{P}_\text{cm}=\text{l}_\text{cm}\times\omega$

$\Rightarrow2\text{m}\times\frac{\text{v}_0}{6}\times\frac{1}{3}+\text{m}\times\frac{\text{v}_0}{3}\times\frac{\text{2l}}{3}$

$=\text{2m}\Big(\frac{1}{3}\Big)^2+\text{m}\Big(\frac{2\text{l}}{3}\Big)^2\times\omega$

$\Rightarrow\frac{6\text{mv}_0\text{l}}{18}=\frac{\text{6ml}}{9}\times\omega$

$\Rightarrow\omega=\frac{\text{v}_0}{\text{2l}}$

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Question 25 Marks

A uniform metre stick of mass 200g is suspended from the ceiling through two vertical strings of equal lengths fixed at the ends. A small object of mass 20g is placed on the stick at a distance of 70cm from the left end. Find the tensions in the two strings.

Answer

According to the question
m₁ = 200g, I = 1m, m₂ = 20g
Therefore, (T₁ × r₁) - (T₂ × r₂) - (m₁f × r₃g) = 0
⇒ T₁ × 0.7 - T₂ × 0.3 - 2 × 0.2 × g = 0
⇒ 7T₁ - 3T₂ = 3.92 ...(1)
T₁ + T₂ = 0.2 × 9.8 + 0.02 × 9.8 = 2.156 ...(2)
From the equation (1) and (2) we will get
10T₁ = 10.3
⇒ T₁ = 1.038N = 1.04N
Therefore, T₂ = 2.156 - 1.038 = 1.118 = 1.12N.

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Question 35 Marks

The door of an almirah is 6ft high, 1.5ft wide and weighs 8kg. The door is supported by two hinges situated at a distance of 1ft from the ends. If the magnitudes of the forces exerted by the hinges on the door are equal, find this magnitude.

Answer

According to the question

8g = F₁ + F₂; N₁ = N₂

Since, R₁ = R₂

Therefore F₁ = F₂

⇒ 2F₁ = 8g

⇒ F₁ = 40

Let us take torque about the point B, we will get N₁ × 4 = 8g × 0.75.

$\Rightarrow\text{N}_1=\frac{(80\times3)}{(4\times4)}=15\text{N}$

Therefore $\sqrt{\Big(\text{F}_1^2+\text{N}_1^2\Big)}=\text{R}_1$

$\sqrt{40^2+15^2}=42.72=43\text{N.}$

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Question 45 Marks

Figure shows a small spherical ball of mass m rolling down the loop track. The ball is released on the linear portion at a vertical height H from the lowest point. The circular part shown has a radius R.

Find the kinetic energy of the ball when it is at a point A where the radius makes an angle $\theta$ with the horizontal.
Find the radial and the tangential accelerations of the centre when the ball is at A.
Find the normal force and the frictional force acting on the ball if H = 60cm, R = 10cm, $\theta=0 $ and m = 70g.

Answer

Total kinetic energy $\text{y}=\Big(\frac{1}{2}\Big)\text{mv}^2+\Big(\frac{1}{2}\Big)\text{l}\omega^2$

Therefore according to the question

$\text{mgH}=\Big(\frac{1}{2}\Big)\text{mv}^2+\Big(\frac{1}{2}\Big)\text{l}\omega^2+\text{mgR}(1+\cos\theta)$

$\Rightarrow\text{mgH}-\text{mgR}(1+\cos\theta)=\Big(\frac{1}{2}\Big)\text{mv}^2+\Big(\frac{1}{2}\Big)\text{l}\omega^2$

$\Rightarrow\Big(\frac{1}{2}\Big)\text{mv}^2+\Big(\frac{1}{2}\Big)\text{l}\omega^2=\text{mg}(\text{H}-\text{R}-\text{R}\sin\theta)$

To find the acceleration components

$\Rightarrow\Big(\frac{1}{2}\Big)\text{mv}^2+\Big(\frac{1}{2}\Big)\text{l}\omega^2=\text{mg}(\text{H}-\text{R}-\text{R}\sin\theta)$

$\Rightarrow\frac{7}{10}\text{mv}^2=\text{mg}(\text{H}-\text{R}-\text{R}\sin\theta)$

$\frac{\text{v}^2}{\text{R}}=\frac{10}{7}\text{g}\Big[\Big(\frac{\text{H}}{\text{R}}\Big)-1-\sin\theta\Big]\rightarrow$ radical acceleration

$\Rightarrow\text{v}^2=\frac{10}{7}\text{g}(\text{H}-\text{R})-\text{R}\sin\theta$

$\Rightarrow\text{2v}\frac{\text{dv}}{\text{dt}}=-\frac{10}{7}\text{g}\text{R}\cos\theta\frac{\text{d}\theta}{\text{dt}}$

$\Rightarrow\omega\text{R}\frac{\text{dv}}{\text{dt}}=-\frac{5}{7}\text{g}\text{R}\cos\theta\frac{\text{d}\theta}{\text{dt}}$

$\Rightarrow\frac{\text{dv}}{\text{dt}}=-\frac{5}{7}\text{g}\cos\theta\rightarrow$ tangential acceleration

Normal force at $\theta=0$

$\Rightarrow\frac{\text{mv}^2}{\text{R}}=\frac{70}{1000}\times\frac{10}{7}\times\Big(\frac{0.6-0.1}{0.1}\Big)=5\text{N}$

Frictional force:

$\text{f}=\text{mg}-\text{ma}=\text{m}(\text{g}-\text{a})$

$=\text{m}\Big(10-\frac{5}{7}\times10\Big)=0.07\Big(\frac{70-50}{7}\Big)$

$=\frac{1}{100}\times20=0.2\text{N}$

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Question 55 Marks

Solve the previous problem if the friction coefficient between the 2.0kg block and the plane below it is 0.5 and the plane below the 4.0kg block is frictionless.

Answer

m₁ = 4kg, m₂ = 2kg

Frictional co-efficient between 2kg block and surface = 0.5

R = 10cm = 0.1m

I = 0.5kg-m²

$\text{m}_1\text{g}\sin\theta-\text{T}_1=\text{m}_1\text{a}\ \dots(1)$

$\text{T}_2-(\text{m}_2\text{g}\sin\theta+\mu\text{m}_2\text{g}\cos\theta)=\text{m}_2\text{a}\ \dots(2)$

$(\text{T}_1-\text{T}_2)=\frac{\text{la}}{\text{r}^2}$

Adding equation (1) and (2) we will get

$\text{m}_1\text{g}\sin\theta-(\text{m}_2\text{g}\sin\theta+\mu\text{m}_2\text{g}\cos\theta)+(\text{T}_1+\text{T}_2)=\text{m}_1\text{a}+\text{m}_2\text{a}$

$\Rightarrow4\times9.8\times\Big(\frac{1}{\sqrt2}\Big)-\Big\{\Big(2\times9.8\times\Big(\frac{1}{\sqrt2}\Big)+0.5\times2\times9.8\times\Big(\frac{1}{\sqrt2}\Big)\Big\}$

$=\Big(4+2+\frac{0.5}{0.01}\Big)\text{a}$

$\Rightarrow27.80-(13.90+6.95)=65\text{a}$

$\Rightarrow\text{a}=0.125\text{ms}^{-2}$

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Question 65 Marks

A uniform rod pivoted at its upper end hangs vertically. It is displaced through an angle of 60° and then released. Find the magnitude of the force acting on a particle of mass dm at the tip of the rod when the rod makes an angle of 37° with the vertical.

Answer

Let l = length of the rod, and m = mass of the rod.

Applying energy principle

$\Big(\frac{1}{2}\Big)\text{l}\omega^2-0=\text{mg}\Big(\frac{1}{2}\Big)(\cos37^\circ-\cos60^\circ)$

$\Rightarrow\frac{1}{2}\times\frac{\text{ml}^2}{3}\omega^2$

$=\text{mg}\times\frac{1}{2}\Big(\frac{4}{5}-\frac{1}{2}\Big)\text{t}$

$\Rightarrow\omega^2=\frac{9\text{g}}{10\text{l}}=0.9\Big(\frac{\text{g}}{\text{l}}\Big)$

Again $\Big(\frac{\text{ml}^2}{3}\Big)\alpha=\text{mg}\Big(\frac{1}{2}\Big)\sin37^\circ=\text{mgl}\times\frac{3}{5}$

$\therefore\alpha=0.9\Big(\frac{\text{g}}{\text{l}}\Big)=$ angular acceleration.

So, to find out the force on the particle at the tip of the rod

F_i = centrifugal force $=(\text{dm})\omega^2\text{l}=0.9(\text{dm})\text{g}$

F_t = tangential force $=(\text{dm})\alpha\text{l}=0.9(\text{dm})\text{g}$

So, total force $\text{F}=\sqrt{\big(\text{F}_{\text{i}}^2+\text{F}_{\text{t}}^2\big)}=0.9\sqrt2(\text{dm})\text{g}$

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Question 75 Marks

A ball is whirled in a circle by attaching it to a fixed point with a string. Is there an angular rotation of the ball about its centre? If yes, is this angular velocity equal to the angular velocity of the ball about the fixed point?

Answer

Yes, there is an angular rotation of the ball about its centre.
Yes, angular velocity of the ball about its centre is same as the angular velocity of the ball about the fixed point.
Explanation:
Let the time period of angular rotation of the ball be T.
Therefore, we get:
Angular velocity of the ball about the fixed point $=\frac{2\pi}{\text{T}}$
After one revolution about the fixed centre is completed, the ball has come back to its original position. In this case, the point at which the ball meets with the string is again visible after one revolution. This means that it has undertaken one complete rotation about its centre.
The ball has taken one complete rotation about its centre. Therefore, we have:
Angular displacement of the ball $=2\pi$
Time period = T
So, angular velocity is again $\frac{2\pi}{\text{T}}.$ Thus, in both the cases, angular velocities are the same.

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Question 85 Marks

Particles of masses 1g, 2g, 3g, ........, 100g are kept at the marks 1cm, 2cm, 3cm, ........, 100cm respectively on a metre scale. Find the moment of inertia of the system of particles about a perpendicular bisector of the metre scale.

Answer

Masses of 1gm, 2gm ........ 100gm are kept at the marks 1cm, 2cm, ......... 1000cm on he x axis respectively. A perpendicular axis is passed at the 50^th particle.

Therefore on the L.H.S. side of the axis there will be 49 particles and on the R.H.S. side there are 50 particles.

Consider the two particles at the position 49cm and 51cm.

Moment inertial due to these two particle will be = 49 × 1² + 51 + 1² = 100gm-cm²

Similarly, if we consider 48^th and 52^nd term we will get 100 × 2²gm-cm²

Therefore we will get 49 such set and one lone particle at 100cm.

Therefore total moment of inertia $=100\big\{1^2+2^2+3^2+\dots+49^2\big\}+100(50)^2$

$=100\times\frac{(50\times51\times101)}{6}=4292500\text{gm}\text{-cm}^2$

$=0.429\text{kg}-\text{m}^2=0.43\text{kg}\text{-m}^2$

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Question 95 Marks

A body is in translational equilibrium under the action of coplanar forces. If the torque of these forces is zero about a point, is it necessary that it will also be zero about any other point?

Answer

Yes, if the torque due to forces in translation equillibriumis zero about a point, it will be zero about other point in the plane.

Let us consider a planner lamina of some mass, acted upon by forces $\vec{\text{F}}_1,\ \vec{\text{F}}_2,\ \dots\ \vec{\text{F}}_{\text{i,}}$ etc.

Let a force $\vec{\text{F}}_1$ act on a i^th particle and torque due to $\vec{\text{F}}_{\text{i}}$ be zeroat a point Q.

Since the body is in translation equillibrium, we have:

$\sum \vec{\text{F}}_{\text{i}}=0\ \dots(1)$

Again, torque about P is zero. Therefore, we have:

$\sum\Big(\vec{\text{r}}_{\text{pi}}\times \vec{\text{F}}_{\text{i}}\Big)=0\ \dots(2)$

Now, torque about point Q will be:

$\sum\vec{\text{r}}_{\text{Qi}}\times \vec{\text{F}}_{\text{i}}$

$=\sum\Big( \overrightarrow{\text{r}}_{\text{QP}}+ \overrightarrow{\text{r}}_{\text{pi}}\Big)\times \overrightarrow{\text{F}}_{\text{i}}$ [From fig.]

$=\sum\Big( \vec{\text{r}}_{\text{Qp}}\times \vec{\text{F}}_{\text{i}}+ \vec{\text{r}}_{\text{pi}}\times \vec{\text{F}}_{\text{i}}\Big)$

$= \overrightarrow{\text{r}_{\text{Qp}}}\times \overrightarrow{\text{F}}_{\text{i}}+ \overrightarrow{\text{r}_{\text{pi}}}\times \overrightarrow{\text{F}}_{\text{i}}$

$=\sum \overrightarrow{\text{r}_{\text{QP}}}\times\sum \overrightarrow{\text{F}}_{\text{i}}+0$ [From (2)]

$ \overrightarrow{\text{r}_{\text{Qp}}}\times0$ [From (1)]

$=0$

Thus, $\overrightarrow{\text{F}}$ is zero about any other point Q.

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Question 105 Marks

A uniform rod of mass m and length l is struck at an end by a force F perpendicular to the rod for a short time interval t. Calculate:

The speed of the centre of mass.
The angular speed of the rod about the centre of mass.
The kinetic energy of the rod.
The angular momentum of the rod about the centre of mass after the force has stopped to act. Assume that t is so small that the rod does not appreciably change its direction while the force acts.

Answer

A uniform rod of mass m length $\ell$ is struck at an end by a force F. $\perp$ to the rod for a short time t:

Speed of the centre of mass:

$\text{mv}=\text{Ft}$

$\Rightarrow\text{v}=\frac{\text{Ft}}{\text{m}}$

The angular speed of the rod about the centre of mass:

$\ell\omega-\text{r}\times\text{p}$

$\Rightarrow\Big(\frac{\text{m}\ell}{12}\Big)\times\omega=\Big(\frac{1}{2}\Big)\times\text{mv}$

$\Rightarrow\Big(\frac{\text{m}\ell}{12}\Big)\times\omega=\Big(\frac{1}{2}\Big)\ell\omega^2$

$\Rightarrow\omega=\frac{\text{m}\ell}{6\text{Ft}}$

$\text{K.E.}=\Big(\frac{1}{2}\Big)\text{mv}^2+\Big(\frac{1}{2}\Big)\ell\omega^2$

$=\Big(\frac{1}{2}\Big)\times\text{m}\Big(\frac{\text{Ft}}{\text{m}}\Big)^2+\Big(\frac{1}{2}\Big)\ell\omega^2$

$=\Big(\frac{1}{2}\Big)\times\text{m}\times\Big(\frac{\text{F}^2\text{t}^2}{\text{m}^2}\Big)+\Big(\frac{1}{2}\Big)\times\Big(\frac{\text{m}\ell^2}{12}\Big)\Big[36\times\Big(\frac{\text{F}^2\text{t}^2}{\text{m}^2\ell^2}\Big)\Big]$

$=\frac{\text{F}^2\text{t}^2}{\text{2m}}+\frac{3}{2}\Big(\frac{\text{F}^2\text{t}^2}{\text{m}}\Big)=2\Big(\frac{\text{F}^2\text{t}^2}{\text{m}}\Big)$

Angular momentum about the centre of mass:

$\text{L}=\text{mvr}=\text{m}\times\frac{\text{Ft}}{\text{m}}\times\Big(\frac{1}{2}\Big)=\frac{\text{F}\ell\text{t}}{2}$

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Question 115 Marks

Calculate the total torque acting on the body shown in figure about the point O.

Answer

Torque about a point = Total force × perpendicular distance from the point to that force.

Let anticlockwise torque = +ve

And clockwise acting torque = -ve

Force acting at the point B is 15N

Therefore torque at O due to this force

$=15\times6\times10^{-2}\times\sin37^\circ$

$=15\times6\times10^{-2}\times\frac{3}{5}=0.54\text{N-m}$ (anticlock wise)

Force acting at the point C is 10N

Therefore, torque at O due to this force

= 10 × 4 × 10^-2 = 0.4N-m (clockwise)

Force acting at the point A is 20N

Therefore, Torque at O due to this force $=20\times4\times10^{-2}\times\sin30^{\circ}$

$=20\times4\times10^{-2}\times\frac{1}{2}=0.4\text{N-m}$ (anticlockwise)

Therefore resultant torque acting at ‘O’ = 0.54 - 0.4 + 0.4 = 0.54N-m.

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Question 125 Marks

The descending pulley shown in figure has a radius 20cm and moment of inertia 0.20kg-m². The fixed pulley is light and the horizontal plane frictionless. Find the acceleration of the block if its mass is 1.0kg.

Answer

A is light pulley and B is the descending pulley having I = 0.20kg-m² and r = 0.2m

Mass of the block = 1kg

According to the equation

$\text{T}_1=\text{m}_1\text{a}\ \dots(1)$

$(\text{T}_2-\text{T}_1)\text{r}=\text{l}\alpha\ \dots(2)$

$\text{m}_2\text{g}-\frac{\text{m}_2\text{a}}{2}=\text{T}_1+\text{T}_2\ \dots(3)$

$\text{T}_2-\text{T}_1=\frac{\text{la}}{2\text{R}^2}=\frac{\text{5a}}2{}$ and $\text{T}_1=\text{a}\ \Big($because $\alpha=\frac{\text{a}}{2\text{R}}\Big)$

$\Rightarrow\text{T}_2=\frac{7}{2}\text{a}$

$\Rightarrow\text{m}_2\text{g}=\frac{\text{m}_2\text{a}}{2}+\frac{7}{2}\text{a}+\text{a}$

$\Rightarrow\frac{2\text{l}}{\text{r}^2\text{g}}=\frac{\frac{\text{2l}}{\text{r}^2}\text{a}}{2}+\frac{9}{2}\text{a}$ $\Big(\frac{1}{2}\text{mr}^2=\text{l}\Big)$

$\Rightarrow98=\text{5a}+4.5\text{a}$

$\Rightarrow\text{a}=\frac{98}{9.5}=10.3\text{ms}^2$

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Question 135 Marks

The pulleys in figure are identical, each having a radius R and moment of inertia I. Find the acceleration of the block M.

Answer

According to the question

$\text{Mg}-\text{T}_1=\text{Ma}\ \dots(1)$

$(\text{T}_1-\text{T}_2)\text{R}=\frac{\text{la}}{\text{R}}$

$\Rightarrow(\text{T}_2-\text{T}_1)=\frac{\text{la}}{\text{R}^2}\ \dots(2)$

$(\text{T}_2-\text{T}_3)\text{R}=\frac{\text{la}}{\text{R}^2}\ \dots(3)$

$\Rightarrow\text{T}_3-\text{mg}=\text{ma}\ \dots(4)$

By adding equation (2) and (3) we will get,

$\Rightarrow(\text{T}_1-\text{T}_3)=2\frac{\text{la}}{\text{R}^2}\ \dots(5)$

By adding equation (1) and (4) we will get,

$-\text{mg}+\text{Mg}+(\text{T}_3-\text{T}_1)=\text{Ma}+\text{ma}\ \dots(6)$

Substituting the value for T₃ - T₁ we will get

$\Rightarrow\text{Mg}-\text{mg}=\text{Ma}+\text{ma}+\frac{2\text{la}}{\text{R}^2}$

$\Rightarrow\text{a}=\frac{(\text{M}-\text{m})\text{G}}{\big(\text{M}+\text{m}+\frac{\text{2l}}{\text{R}^2}\big)}$

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Question 145 Marks

A metre stick weighing 240g is pivoted at its upper end in such a way that it can freely rotate in a vertical plane through this end (figure). A particle of mass 100g is attached to the upper end of the stick through a light string of length 1m. Initially, the rod is kept vertical and the string horizontal when the system is released from rest. The particle collides with the lower end of the stick and sticks there. Find the maximum angle through which the stick will rise.

Answer

$\frac{1}{2}\text{l}\omega^2-0=0.1\times10\times1$

$\Rightarrow\omega=\sqrt{20}$

For collision

$0.1\times1^2\times\sqrt{20}+0=\Big[\Big(\frac{0.24}{3}\Big)\times1^2+(0.1)^21^2\Big]\omega'$

$\Rightarrow\omega'=\frac{\sqrt{20}}{10.(0.18)}$

$\Rightarrow0-\frac{1}{2}\omega'^2=-\text{m}_1\text{gl}(1-\cos\theta)-\text{m}_2\text{g}\frac{\text{l}}2{}(1-\cos\theta)$

$=0.1\times10(1-\cos\theta)=0.24\times10\times0.5(1-\cos\theta)$

$\Rightarrow\frac{1}{2}\times0.18\times\Big(\frac{20}{3.24}\Big)=2.2(1-\cos\theta)$

$\Rightarrow(1-\cos\theta)=\frac{1}{(2.2\times1.8)}$

$\Rightarrow1-\cos\theta=0.252$

$\Rightarrow\cos\theta=1-0.252=0.748$

$\Rightarrow\omega=\cos^{-1}(0.748)=41^\circ$

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Question 155 Marks

Suppose the friction coefficient between the ground and the ladder of the previous problem is 0.540. Find the maximum weight of a mechanic who could go up and do the work from the same position of the ladder.

Answer

$\mu=0.54,\ \text{R}_2=16\text{g}+\text{mg};\ \text{R}_1=\mu\text{R}_2$

$\Rightarrow\text{R}_1\times10\cos37^\circ=16\text{g}\times5\sin37^\circ+\text{mg}\times8\times\sin37^\circ$

$\Rightarrow8\text{R}_1=48\text{g}+\frac{24}{5}\text{mg}$

$\Rightarrow\text{R}_1=\frac{48\text{g}+\frac{24}{5}\text{mg}}{8\times0.54}$

$\Rightarrow16\text{g}+\text{mg}=\frac{24.0\text{g}+24\text{mg}}{5\times8\times0.54}$

$\Rightarrow16+\text{m}=\frac{240+24\text{m}}{40\times0.54}$

$\Rightarrow\text{m}=44\text{kg}.$

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Question 165 Marks

A uniform rod of length L lies on a smooth horizontal table. A particle moving on the table strikes the rod perpendicularly at an end and stops. Find the distance travelled by the centre of the rod by the time it turns through a right angle. Show that if the mass of the rod is four times that of the particle, the collision is elastic.

Answer

Let the mass of the particle = m & the mass of the rod = M.

Let the particle strikes the rod with a velocity V.

If we take the two body to be a system.

Therefore the net external torque & net external force = 0

Therefore Applying laws of conservation of linear momentum MV' = mV (V' = velocity of the rod after striking)

$\Rightarrow\frac{\text{V}'}{\text{V}}=\frac{\text{m}}{\text{M}}$

Again applying laws of conservation of angular momentum

$\Rightarrow\frac{\text{mVR}}{2}=\ell\omega$

$\Rightarrow\frac{\text{mVR}}{2}=\frac{\text{MR}^2}{12}\times\frac{\pi}{2\text{t}}$

$\Rightarrow\text{t}=\frac{\text{MR}\pi}{\text{m}\times12\times\text{V}}$

Therefore distance travelled:

$\text{V}'\text{t}=\text{V}'\Big(\frac{\text{MR}\pi}{\text{m}\times12\times\pi}\Big)$

$=\frac{\text{m}}{\text{M}}\times\frac{\text{M}}{\text{m}}\times\frac{\text{R}\pi}{12}=\frac{\text{R}\pi}{12}$

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Question 175 Marks

A 6.5m long ladder rests against a vertical wall reaching a height of 6.0m. A 60kg man stands half way up the ladder.

Find the torque of the force exerted by the man on the ladder about the upper end of the ladder.
Assuming the weight of the ladder to be negligible as compared to the man and assuming the wall to be smooth, find the force exerted by the ground on the ladder.

Answer

m = 60kg, ladder length = 6.5m, height of the wall = 6m

Therefore torque due to the weight of the body

$\tau=\frac{600\times6.5}{2\sin\theta}=\text{i}$

$\Rightarrow\tau=\frac{600\times6.5}{2\times\sqrt{\big[1-\big(\frac{6}{6.5}\big)^2\big]}}$

$\Rightarrow\tau=735\text{N-m}$

$\text{R}_2=\text{mg}=60\times9.8$

$\text{R}_1=\mu\text{R}_2$

$\Rightarrow6.5\text{R}_1\cos\theta=60\text{g}\sin\theta\times\frac{6.5}{2}$

$\Rightarrow\text{R}_1=60\text{g}\tan\theta=60\text{g}\times\Big(\frac{2.5}{12}\Big)\ \Big[$Because $\tan\theta=\frac{2.5}{6}\Big]$

$\Rightarrow\text{R}_1=\Big(\frac{25}{2}\Big)\text{g}=122.5\text{N}$

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Question 185 Marks

A uniform ladder of length 10.0m and mass 16.0kg is resting against a vertical wall making an angle of 37° with it. The vertical wall is frictionless but the ground is rough. An electrician weighing 60.0kg climbs up the ladder. If he stays on the ladder at a point 8.00m from the lower end, what will be the normal force and the force of friction on the ladder by the ground? What should be the minimum coefficient of friction for the elctrician to work safely?

Answer

$\text{R}_1=\mu\text{R}_2,\ \text{R}_2=16\text{g}+60\text{g}=745\text{N}$

$\text{R}_1\times10\cos37^\circ=16\text{g}\times5\sin37^\circ+60\text{g}\times8\times\sin37^\circ$

$\Rightarrow8\text{R}_1=\text{48g}+288\text{g}$

$\Rightarrow\text{R}_1=\frac{336\text{g}}{8}=412\text{N}=\text{f}$

Therefore $\mu=\frac{\text{R}_1}{\text{R}_2}=\frac{412}{745}=0.53$

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Question 195 Marks

Two blocks of masses 400g and 200g are connected through a light string going over a pulley which is free to rotate about its axis. The pulley has a moment of inertia 1.6 x 10^-4kg-m² and a radius 2.0cm. Find:

The kinetic energy of the system as the 400g block falls through 50cm.
The speed of the blocks at this instant.

Answer

Total kinetic energy of the system

According to the question

$0.4\text{g}-\text{T}_1=0.4\text{a}\ \dots(1)$

$\text{T}_2-0.2\text{g}=0.2\text{a}\ \dots(2)$

$(\text{T}_1-\text{T}_2)\text{r}=\frac{\text{la}}{\text{r}}\ \dots(3)$

From equation 1, 2 and 3

$\Rightarrow\text{a}=\frac{(0.4-0.2)\text{g}}{\big(0.4+0.2+\frac{1.6}{0.4}\big)}=\frac{\text{g}}5{}$

Therefore,

$\text{V}=\sqrt{2\text{ah}}=\sqrt{(2\times\text{gl}^5\times0.5)}$

$\Rightarrow\sqrt{\Big(\frac{\text{g}}5{}\Big)}=\sqrt{\Big(\frac{\text{9.8}}5{}\Big)}=1.4\text{m/s}.$

$\frac{1}{2}\text{m}_1\text{V}^2+\frac{1}{2}\text{m}_2\text{V}^2+\frac{1}{2}18^2$

$=\Big(\frac{1}{2}\times0.4\times1.4^2\Big)+\Big(\frac{1}{2}\times0.2\times1.4^2\Big)\\+\Big(\frac{1}{2}\times\Big(\frac{1.6}{4}\Big)\times1.4^2\Big)$

$=0.98\ \text{Joule.}$

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Question 205 Marks

Suppose the rod with the balls A and B of the previous problem is clamped at the centre in such a way that it can rotate freely about a horizontal axis through the clamp. The system is kept at rest in the horizontal position. A particle P of the same mass m is dropped from a height h on the ball B. The particle collides with B and sticks to it:

Find the angular momentum and the angular speed of the system just after the collision.
What should be the minimum value of h so that the system makes a full rotation after the collision.

Answer

The system is kept rest in the horizontal position and a particle P falls from a height h and collides with B and sticks to it.

Therefore, the velocity of the particle ‘P’ before collision $=\sqrt{2\text{gh}}$

If we consider the two bodies P and B to be a system. Net external torque and force = 0

Therefore, $\text{m}\sqrt{2\text{gh}}=\text{2m}\times\text{v}$

$\Rightarrow\text{v}'=\sqrt{\frac{(2\text{gh})}{2}}$

Therefore angular momentum of the rod just after the collision

$\Rightarrow\text{2m}(\text{v}'\times\text{r})=\text{2m}\times\sqrt{\frac{(2\text{gh})}{2}}\times\frac{\text{l}}2{}$

$\Rightarrow\text{ml}\sqrt{\frac{(2\text{gh})}{2}}$

$\omega=\frac{\text{L}}{\text{l}}=\frac{\text{ml}\sqrt{2\text{gh}}}{2\big(\frac{\text{ml}^2}{4}+\frac{2\text{ml}^2}{4}\big)}$

$=\frac{2\sqrt{\text{gh}}}{\text{3l}}=\frac{\sqrt{8\text{gh}}}{\text{3l}}$

When the mass 2m will at the top most position and the mass m at the lowest point, they will automatically rotate. In this position the total gain in potential energy $=2\text{mg}\times\Big(\frac{\text{l}}{2}\Big)-\text{mg}\Big(\frac{\text{l}}{2}\Big)=\text{mg}\Big(\frac{\text{l}}{2}\Big)$

Therefore

$\text{mg}\frac{\text{l}}{2}=\frac{\text{l}}{2}\text{l}\omega^2$

$\Rightarrow\text{mg}\frac{\text{l}}{2}=\frac{\big(\frac{1}{2}\times3\text{ml}^2\big)}{4}\times\Big(\frac{\text{8gh}}{9\text{gl}^2}\Big)$

$\Rightarrow\text{h}=\frac{3\text{l}}{2}$

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Question 215 Marks

A uniform rod of length L rests against a smooth roller as shown in figure. Find the friction coefficient between the ground and the lower end if the minimum angle that the rod can make with the horizontal is $\theta.$

Answer

Rod has a length = L

It makes an angle $\theta$ with the floor

The vertical wall has a height = h

$\text{R}_2=\text{mg}-\text{R}_1\cos\theta\ \dots(1)$

$\text{R}_1\sin\theta=\mu\text{R}_2\ \dots(2)$

$\text{R}_1\cos\theta\times\Big(\frac{\text{h}}{\tan\theta}\Big)+\text{R}_1\sin\theta\times\text{h}=\text{mg}\times\frac{1}{2}\cos\theta$

$\Rightarrow\text{R}_1\times\Big(\frac{\cos^2\theta}{\tan\theta}\Big)+\text{R}_1\sin\theta\times\text{h}=\text{mg}\times\frac{1}{2}\cos\theta$

$\Rightarrow\text{R}_1=\frac{\text{mg}\times\frac{\text{L}}{2\cos\theta}}{\Big\{\Big(\frac{\cos^2\theta}{\sin\theta}\Big)\text{h}+\sin\theta\text{h}\Big\}}$

$\Rightarrow\text{R}_1\cos\theta=\frac{\frac{\text{mg}\text{L}}{2\cos^2\theta\sin\theta}}{\Big\{\Big(\frac{\cos^2\theta}{\sin\theta}\Big)\text{h}+\sin\theta\text{h}\Big\}}$

$\Rightarrow\mu=\frac{\text{R}_1\sin\theta}{\text{R}_2}=\frac{\frac{\text{mgL}}{2\cos\theta\sin\theta}}{\Big\{\Big(\frac{\cos^2\theta}{\sin\theta}\Big)\text{h}+\sin\theta\text{h}\Big\}\text{mg}-\text{mg}\frac{1}{2}\cos^2\theta}$

$\Rightarrow\mu=\frac{\frac{\text{L}}{2\cos\theta.\sin\theta}\times2\sin\theta}{2\big(\cos^2\theta\text{h}+\sin^2\theta\text{h}\big)-\text{L}\cos^2\theta\sin\theta}$

$\Rightarrow\mu=\frac{\text{L}\cos\theta\sin^2\theta}{\text{2h}-\text{L}\cos^2\theta\sin\theta}$

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Question 225 Marks

The pulley shown in figure has a radius 10cm and moment of inertia 0.5kg-m² about its axis. Assuming the inclined planes to be frictionless, calculate the acceleration of the 4.0kg block.

Answer

$\text{m}_1\text{g}\sin\theta-\text{T}_1=\text{m}_1\text{a}\ \dots(1)$

$(\text{T}_1-\text{T}_2)=\frac{\text{la}}{\text{r}^2}\ \dots(2)$

$\text{T}_2-\text{m}_2\text{g}\sin\theta=\text{m}_2\text{a}\ \dots(3)$

Adding the equations (1) and (3) we will get

$\text{m}_1\text{g}\sin\theta+(\text{T}_2-\text{T}_1)-\text{m}_2\text{g}\sin\theta=(\text{m}_1+\text{m}_2)\text{a}$

$\Rightarrow(\text{m}_1-\text{m}_2)\text{g}\sin\theta=\Big(\text{m}_1+\text{m}_2+\frac{1}{\text{r}^2}\Big)\text{a}$

$\Rightarrow\text{a}=\frac{(\text{m}_1-\text{m}_2)\text{g}\sin\theta}{\Big(\text{m}_1+\text{m}_2+\frac{1}{\text{r}^2}\Big)}=0.248=0.25\text{ms}^{-2}$

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Question 235 Marks

Suppose the smaller pulley of the previous problem has its radius 5.0cm and moment of inertia 0.10kg-m². Find the tension in the part of the string joining the pulleys.

Answer

m = 2kg, i₁ = 0.10kg-m², r₁ = 5cm = 0.05m

i₂ = 0.20kg-m², r₂ = 10cm = 0.1m

Therefore $\text{mg}-\text{T}_1=\text{ma}\ \dots(1)$

$\big(\text{T}_1-\text{T}_2\big)\text{r}_1=\text{l}_1\alpha\ \dots(2)$

$\text{T}_2\text{r}_2=\text{l}_2\alpha\ \dots(3)$

Substituting the value of T₂ in the equation (2), we get

$\Rightarrow\Big(\text{t}_1-\frac{\text{l}_2\alpha}{\text{r}_1}\Big)\text{r}_2=\text{l}_1\alpha$

$\Rightarrow\Big(\text{T}_1-\frac{\text{l}_2\text{a}}{\text{r}_1^2}\Big)=\frac{\text{l}_1\text{a}}{\text{r}_2^2}$

$\Rightarrow\text{T}_1\Big[\Big(\frac{\text{l}_1}{\text{r}_1^2}\Big)+\Big(\frac{\text{l}_2}{\text{r}_2^2}\Big)\Big]\text{a}$

Substituting the value of T₁ in the equation (1), we get

$\Rightarrow\text{mg}-\Big[\Big(\frac{\text{l}_1}{\text{r}_1^2}\Big)+\Big(\frac{\text{l}_2}{\text{r}_2^2}\Big)\Big]\text{a}=\text{ma}$

$\Rightarrow\frac{\text{mg}}{\Big[\Big(\frac{l_1}{\text{r}_1^2}\Big)+\Big(\frac{l_2}{\text{r}_2^2}\Big)\Big]+\text{m}}=\text{a}$

$\Rightarrow\text{a}=\frac{2\times9.8}{\Big(\frac{0.1}{0.0025}\Big)+\Big(\frac{0.2}{0.01}\Big)+2}=0.316\text{m/s}^2$

$\Rightarrow\text{T}_2=\frac{\text{l}_2\text{a}}{\text{r}_2^2}=\frac{0.20\times0.316}{0.01}=6.32\text{N}$

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Question 245 Marks

A uniform wheel of radius R is set into rotation about its axis at an angular speed $\omega.$ This rotating wheel is now placed on a rough horizontal surface with its axis horizontal. Because of friction at the contact, the wheel accelerates forward and its rotation decelerates till the wheel starts pure rolling on the surface. Find the linear speed of the wheel after it starts pure rolling.

Answer

A uniform wheel of radius R is set into rotation about its axis (case-I) at an angular speed $\omega.$

This rotating wheel is now placed on a rough horizontal. Because of its friction at contact, the wheel accelerates forward and its rotation decelerates. As the rotation decelerates the frictional force will act backward.

If we consider the net moment at A then it is zero.

Therefore the net angular momentum before pure rolling & after pure rolling remains constant

Before rolling the wheel was only rotating around its axis.

Therefore Angular momentum $=\ell\omega=\Big(\frac{1}{2}\Big)\text{MR}^2\omega\ \dots(1)$

After pure rolling the velocity of the wheel let v

Therefore angular momentum $=\ell_{\text{cm}}\omega+\text{m}(\text{v}\times\text{R})$

$=\Big(\frac{1}{2}\Big)\text{m}\text{R}^2\Big(\frac{\text{v}}{\text{R}}\Big)+\text{mvR}=\frac{3}{2}\text{mvR}\ \dots(2)$

Because, Eq. (1) and (2) are equal

Therefore, $\frac{3}{2}\text{mvR}=\frac{1}{2}\text{mR}^2\omega$

$\Rightarrow\text{V}=\omega\frac{\text{R}}{3}$

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Question 255 Marks

A solid sphere of mass 0.50kg is kept on a horizontal surface. The coefficient of static friction between the surfaces in contact is $\frac{2}{7}.$ What maximum force can be applied at the highest point in the horizontal direction so that the sphere does not slip on the surface?

Answer

If we take moment about the centre, then

$\text{F}\times\text{R}=\ell\alpha\times\text{f}\times\text{R}$

$\Rightarrow\text{F}=\frac{2}{5}\text{mR}\alpha+\mu\text{mg}\ \dots(1)$

Again, $\text{F}=\text{ma}_{\text{c}}-\mu\text{mg}\ \dots(2)$

$\Rightarrow\text{a}_\text{c}=\frac{\text{F}+\mu\text{mg}}{\text{m}}$

Putting the value a_c in eq(1) we get

$\Rightarrow\frac{2}{5}\times\text{m}\times\Big(\frac{\text{F}+\mu\text{mg}}{\text{m}}\Big)+\mu\text{mg}$

$\Rightarrow\frac{2}{5}(\text{F}+\mu\text{mg})+\mu\text{mg}$

$\Rightarrow\text{F}=\frac{2}{5}\text{F}+\frac{2}{5}\times0.5\times10+\frac{2}{7}\times0.5\times10$

$\Rightarrow\frac{\text{3F}}{5}=\frac{4}{7}+\frac{10}{7}=2$

$\Rightarrow\text{F}=\frac{5\times2}{3}=\frac{10}{3}=3.33\text{N}$

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Question 265 Marks

A hollow sphere is released from the top of an inclined plane of inclination $\theta.$

What should be the minimum coefficient of friction between the sphere and the plane to prevent sliding?
Find the kinetic energy of the ball as it moves down a length 1 on the incline if the friction coefficient is half the value calculated in part (a).

Answer

A hollow sphere is released from a top of an inclined plane of inclination $\theta.$ To prevent sliding, the body will make only perfect rolling. In this condition,

$\text{mg}\sin\theta-\text{f}=\text{ma}\ \dots(1)$

& torque about the centre

$\text{f}\times\text{R}=\frac{2}{3}\text{mR}^2\times\frac{\text{a}}{\text{R}}$

$\Rightarrow\text{f}=\frac{2}{3}\text{ma}\ \dots(2)$

Putting this value in equation (1) we get

$\Rightarrow\text{mg}\sin\theta-\frac{2}{3}\text{ma}=\text{ma}$

$\Rightarrow \text{a}=\frac{3}{5}\text{g}\sin\theta$

$\Rightarrow\text{mg}\sin\theta-\text{f}=\frac{3}{5}\text{mg}\sin\theta$

$\Rightarrow\text{f}=\frac{2}{5}\text{mg}\sin\theta$

$\Rightarrow\mu\text{mg}\cos\theta=\frac{2}{5}\text{mg}\sin\theta$

$\mu=\frac{2}{5}\tan\theta$

$\frac{1}{5}\tan\theta(\text{mg}\cos\theta)\text{R}=\frac{2}{3}\text{mR}^2\alpha$

$\Rightarrow\alpha=\frac{3}{10}\times\frac{\text{g}\sin\theta}{\text{R}}$

$\text{a}_{\text{c}}=\text{g}\sin\theta-\frac{\text{g}}{5}\sin\theta=\frac{4}{5}\sin\theta$

$\Rightarrow\text{t}^2=\frac{\text{2s}}{\text{a}_{\text{c}}}=\frac{\text{2l}}{\big(\frac{4\text{g}\sin\theta}{5}\big)}=\frac{5\text{l}}{2\text{g}\sin\theta}$

Again, $\omega=\alpha\text{t}$

$\text{K.E.}=\Big(\frac{1}{2}\Big)\text{mv}^2+\Big(\frac{1}{2}\Big)\text{l}\omega^2$

$=\Big(\frac{1}{2}\Big)\text{m}(\text{2as})+\Big(\frac{1}{2}\Big)\text{l}\big(\alpha^2\text{t}^2\big)$

$=\frac{1}{2}\text{m}\times\frac{4\text{g}\sin\theta}{5}\times2\times\text{l}+\frac{1}{2}\\\times\frac{2}{3}\text{mR}^2\times\frac{9}{100}\frac{\text{g}^2\sin^2\theta}{\text{R}}\times\frac{\text{5l}}{2\text{g}\sin\theta}$

$=\frac{\text{4mgl}\sin\theta}{5}+\frac{\text{3mgl}\sin\theta}{40}=\frac{7}{8}\text{mgl}\sin\theta$

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Question 275 Marks

Suppose the particle of the previous problem has a mass m and a speed v before the collision and it sticks to the rod after the collision. The rod has a mass M:

Find the velocity of the centre of mass C of the system constituting ''The rod plus the particle''.
Find the velocity of the particle with respect to C before the collision.
Find the velocity of the rod with respect to C before the collision.
Find the angular momentum of the particle and of the rod about the centre of mass C before the collision.
Find the moment of inertia of the system about the vertical axis through the centre of mass C after the collision.
Find the velocity of the centre of mass C and the angular velocity of the system about the centre of mass after the collision.

Answer

If we take the two bodies as a system therefore total external force = 0

Applying L.C.L.M:

$\text{mV}=(\text{M}+\text{m})\text{v}'$

$\Rightarrow\text{v}'=\frac{\text{m}\text{v}}{\text{M}+\text{m}}$

Let the velocity of the particle w.r.t. the centre of mass = V'

$\Rightarrow\text{v}'=\frac{\text{m}\times0+\text{mv}}{\text{M}+\text{m}}$

$\Rightarrow\text{v}'=\frac{\text{Mv}}{\text{M}+\text{m}}$

If the body moves towards the rod with a velocity of v, i.e. the rod is moving with a velocity - v towards the particle.

Therefore the velocity of the rod w.r.t. the centre of mass = V^-

$\Rightarrow\text{V}^{-}=\frac{\text{M}\times\text{O}=\text{m}\times\text{v}}{\text{M}+\text{m}}=\frac{-\text{mv}}{\text{M}+\text{m}}$

The distance of the centre of mass from the particle

$=\frac{\text{M}\times\frac{1}{2}+\text{m}\times\text{O}}{(\text{M}+\text{m})}=\frac{\text{M}\times\frac{1}{2}}{(\text{M}+\text{m})}$

Therefore angular momentum of the particle before the collision

$=\text{l}\omega=\text{Mr}^2\text{cm}\omega$

$=\text{m}\Bigg\{\frac{\text{m}\big(\frac{1}{2}\big)}{(\text{M}+\text{m})}\Bigg\}^2\times\frac{\text{V}}{\frac{1}{2}}$

$=\frac{(\text{mM}^2\text{vl})}{2(\text{M}+\text{m})}$

Distance of the centre of mass from the centre of mass of the rod $=\text{R}^1_{\text{cm}}=\frac{\text{M}\times0+\text{m}\times\big(\frac{1}{2}\big)}{(\text{M}+\text{m})}=\frac{\big(\frac{\text{ml}}{2}\big)}{(\text{M}+\text{m})}$

Therefore angular momentum of the rod about the centre of mass

$=\text{MV}_{\text{cm}}\text{R}_{\text{cm}}^1$

$=\text{M}\times\Big\{\frac{(-\text{mv})}{(\text{M}+\text{m})}\Big\}\Bigg\{\frac{\big(\frac{\text{ml}}2{}\big)}{(\text{M}+\text{m})}\Bigg\}$

$=\Bigg|\frac{-\text{Mm}^2\text{lv}}{2(\text{M}+\text{m})^2}\Bigg|=\frac{\text{Mm}^2\text{lv}}{2(\text{M}+\text{m})^2}$ (If we consider the magnitude only)

Moment of inertia of the system = M.I. due to rod + M.I. due to particle

$=\frac{\text{Ml}^2}{12}+\frac{\text{M}\big(\frac{\text{ml}}{2}\big)^2}{(\text{M}+\text{m})^2}+\frac{\text{m}\big(\frac{\text{Ml}}{\text{S}}\big)^2}{(\text{M}+\text{m})^2}$

$=\frac{\text{Ml}^2(\text{M}+\text{4m})}{12(\text{M}+\text{m})}$

Velocity of the centre of mass $\text{V}_{\text{m}}=\frac{\text{M}\times0+\text{mV}}{(\text{M}+\text{m})}=\frac{\text{mV}}{(\text{M}+\text{m})}$

(Velocity of centre of mass of the system before the collision = Velocity of centre of mass of the system after the collision)

(Because External force = 0)

Angular velocity of the system about the centre of mass,

$\text{P}_{\text{cm}}=\text{l}_{\text{cm}}\omega$

$\Rightarrow\text{M}\vec{\text{V}}_{\text{M}}\times\vec{\text{r}}_{\text{m}}+\text{m}\vec{\text{v}}_{\text{m}}\times\vec{\text{r}}_{\text{m}}=\text{l}_{\text{cm}}\omega$

$\Rightarrow\text{M}\times\frac{\text{mv}}{(\text{M}+\text{m})}\times\frac{\text{ml}}{2(\text{M}+\text{m})}+\text{m}\times\frac{\text{Mv}}{(\text{M}+\text{m})}\times\frac{\text{Ml}}{2(\text{M}+\text{m})}$

$=\frac{\text{Ml}^2(\text{M}+\text{4m})}{12(\text{M}+\text{m})}\times\omega$

$\Rightarrow\frac{\text{M}\text{m}^2\text{vl}+\text{m}\text{M}^2\text{vl}}{2(\text{M}+\text{m})^2}=\frac{\text{Ml}^2(\text{M}+\text{m})}{12(\text{M}+\text{m})}\times\omega$

$\Rightarrow\frac{\frac{\text{Mm}}{(\text{M}+\text{m})}}{2(\text{M}+\text{m})^2}=\frac{\text{Ml}^2(\text{M}+\text{m})}{12(\text{M}+\text{m})}\times\omega$

$\Rightarrow\frac{6\text{mv}}{(\text{M}+\text{4m})\text{l}}=\omega$

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