Work Energy and Power Test

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Work Energy and Power Test - 66

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Question 1
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The simple $$2 kg$$ pendulum is released from rest in the horizontal position. As it reaches the bottom position, the cord wraps around the smooth fixed pin at $$B$$ and continues in the smaller are in the vertical plane. Calculate the magnitude of the force $$R$$ supported by the pin at $$B$$ when the pendulum passes the position $$\displaystyle \theta = 30^{\circ}.\left ( g= 9.8m/s^{2} \right )$$

A
$$15 N$$

B
$$30 N$$

C
$$45 N$$

D
$$60 N$$

Solution

using energy conservation
$$\Rightarrow ({ K.E.) }_{ E }+({ P.E.) }_{ E }=({ K.E.) }_{ C }+({ P.E.) }_{ C }$$
$$\Rightarrow 0+2g\times 800\times { 10 }^{ -3 }=\frac { 1 }{ 2 } \times 2\times { { v }_{ 1 }^{ 2 } }+0$$
$$\Rightarrow v_{ 1 }=4m/s$$
again using energy conservation at point $$C$$ and $$D$$
$$\Rightarrow ({ K.E.) }_{ C }+({ P.E.) }_{ C }=({ K.E.) }_{ D }+({ P.E.) }_{ D }$$
$$\Rightarrow \frac { 1 }{ 2 } \times 2\times { { v }_{ 1 }^{ 2 } }+0=\frac { 1 }{ 2 } \times 2\times { { v }_{ 2 }^{ 2 }+ }2g \times 400(1-\cos { 30° } )\times { 10 }^{ -3 }$$
$$\Rightarrow { v }_{ 2 }^{ 2 }=8+4\sqrt { 3 } $$
From the diagram, at point $$D$$
$$ \Rightarrow T=mg\cos { \theta } +\frac { m{ v }_{ 2 }^{ 2 } }{ R } \\ \Rightarrow T=2\times 10\times \cos { 30° } +\frac { 2\times (8+\sqrt { 3 } ) }{ 400\times { 10 }^{ -3 } } $$
$$\Rightarrow T=90.2N$$
Question 2
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The bob of the pendulum shown in figure describes an arc of circle in a vertical plane. If the tension in the cord is $$2.5\ times$$ the weight of the bob for the position shown. Find the velocity and the acceleration of the bob in that position.

A
$$\displaystyle 16.75ms^{-1}, 5.66ms^{-2}$$

B
$$\displaystyle 5.66ms^{-1}, 16.75ms^{-2}$$

C
$$\displaystyle 2.88ms^{-1}, 16.75ms^{-2}$$

D
$$\displaystyle 5.66ms^{-1}, 8.34ms^{-2}$$

Solution

Let the mass of the bob be $$m$$ and the velocity of the at point B be $$v$$
Given : $$r =2$$ m $$T =2.5 mg$$
Using circular motion equation at B : $$\dfrac{mv^2}{r} = T - mg cos30^o$$
$$\therefore$$ $$\dfrac{mv^2}{2} = 2.5 mg - mg \times 0.866$$
OR $$v^2 = 3.27 g =3.27 \times 9.8 = 32 $$
$$\implies v =5.66 $$ m/s
$$\therefore$$ Radial acceleration $$a_r = \dfrac{v^2}{r} = \dfrac{32}{2} =16$$ $$m/s^2$$
Also tangential acceleration $$a_t = g sin 30^o =9.8 \times 0.5 = 4.9$$ $$m/s^2$$
Total acceleration $$a = \sqrt{a_r^2 + a_t^2} = \sqrt{16^2 + (4.9)^2} = 16.75$$ $$m/s^2$$
Question 3
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The mass of the bob of a simple pendulum of length $$L$$ is $$m$$. If the bob is left from its horizontal position then the speedof the bob and the tension in the thread in the lowest position of the bob will be respectively:

A
$$\;\sqrt{2gL}\;and\;3\;mg$$

B
$$\;3\;mg\;and \sqrt{2gL}$$

C
$$\;2\;mg\;and\;\sqrt{2gL}$$

D
$$\;2\;gL\;and\;3\;mg$$

Solution

Velocity of the bob at point A is zero, thus its kinetic energy at A is zero.
Let the velocity of the bob at point B be $$v$$ and tension in the string be $$T$$.
Using work-energy theorem from A to B : $$W = \Delta K.E$$
$$\therefore $$ $$mgL = \dfrac{1}{2} mv^2 - 0$$ $$\implies v =\sqrt{ 2gL}$$

Circular motion equation at B : $$\dfrac{mv^2}{L} = T - mg$$
$$\therefore$$ $$T = \dfrac{mv^2}{L} + mg = \dfrac{m (2gL)}{L} + mg = 3mg$$
Question 4
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A weightless thread can withstand tension upto $$30\;N$$. A stone of mass $$0.5\ kg$$ is tied to it and is revolved in a circular path of radius $$2\ m$$ in a vertical plane. If $$g=10\ m/s^2$$, then the maximum angular velocity of the stone will be

A
$$5\ rad/s$$

B
$$\sqrt{30}\ rad/s$$

C
$$\sqrt{60}\ rad/s$$

D
$$10\ rad/s$$

Solution

Given : $$T_{max} =30$$ N $$m = 0.5 kg$$ $$r =2$$ m $$g = 10 m/s^2$$
Tension in the string is maximum when the stone is at the lowest position of the vertical motion.
Let the maximum angular velocity be $$w_{max}$$.
Using circular motion equation : $$mrw^2_{max} =T_{max} - mg$$
$$\therefore$$ $$0.5 \times 2 \times w^2_{max} =30 - 0.5 \times 10$$
OR $$w^2_{max} = 30 - 5 =25$$ $$\implies w_{max} = 5$$ $$rad/s$$
Question 5
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A weightless thread can bear tension upto $$3.7\;kg$$ wt. A stone of mass $$500\;gm$$ is tied to it and revolved in a circular path of radius $$4m$$ in a vertical plane. If $$g=10\;ms^{-2}$$, then the maximum angular velocity of the stone will be:

A
$$\;16\;rad/s$$

B
$$\;\sqrt{21}\;rad/s$$

C
$$\;2\;rad/s$$

D
$$\;4\;rad/s$$

Solution

Given : $$T_{max} =3.7 kg wt = 37$$ N $$m =500$$ gm $$= 0.5 kg$$ $$r =4$$ m $$g = 10 m/s^2$$
Tension in the string is maximum when the stone is at the lowest position of the vertical motion.
Let the maximum angular velocity be $$w_{max}$$.
Using circular motion equation : $$mrw^2_{max} =T_{max} - mg$$
$$\therefore$$ $$0.5 \times 4 \times w^2_{max} =37 - 0.5 \times 10$$
OR $$w^2_{max} =16$$ $$\implies w_{max} = 4$$ $$rad/s$$
Question 6
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A weightless thread can bear tension upto $$37 N$$. A stone of mass $$500 g$$ is tied to it and revolved in a circular path of radius $$4 m$$ in a vertical plane. If $$g=10 {ms}^{-2}$$, then the maximum angular velocity of the stone will be:

A
$$2\ rad\ {s}^{-1}$$

B
$$4\ rad\ {s}^{-1}$$

C
$$8\ rad\ {s}^{-1}$$

D
$$16\ rad\ {s}^{-1}$$

Solution

Maximum tension in the string will be at the lowest point of the circular path, where the tension will support the weight as well as provide the centripetal force for circular motion.

So, we have:

$$T=mg+ml{ \omega }^{ 2 }$$

$$\Rightarrow ml{ \omega }^{ 2 }=T-5$$

For maximum angular speed, the tension must be $$37N$$

So, the maximum angular speed is $$\sqrt { \dfrac { T-5 }{ ml } } =\sqrt { \dfrac { 37-5 }{ 0.5\times 4 } } = 4 rad/s$$

Thus, $${ \omega }_{ max }= 4 rad/s $$
Question 7
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Stone tied at one end of light string is whirled round a vertical circle. If the difference between the maximum and minimum tension experienced by the string wire is $$2\ kg\ wt$$, then the mass of the stone must be

A
$$1\ kg$$

B
$$6\ kg$$

C
$$1/3\ kg$$

D
$$2\ kg$$

Solution

$$\textbf{Step 1: FBD [Ref. Fig.]}$$
Tension will be maximum at bottom and minimum at top

$$\textbf{Step 2: Velocities at highest and lowest points}$$
Minimum velocity required at bottom point to complete circle $$= v$$
$$v = \sqrt{5gR}$$ $$....(1)$$

$$\textbf{Step 3: Calculations }$$
For revolving in complete circle tension at highest point $$T\geq 0$$
$$\Rightarrow T_{\min} = 0$$

Applying Newton's second law on the stone along centripetal direction (Considering direction towards center as positive)
$$\Sigma F_c = ma_c$$ $$=m\left (\cfrac{v^2}{R} \right)$$
Using equation $$(1)$$ in the following

$$\Rightarrow T_{max} - mg = \dfrac{mv^2}{R} = 5 mg$$

$$\Rightarrow T_{\max} = 6 mg$$ [T is maximum at bottom]

So, $$ T_{\max} - T_{\min} = 6mg$$
$$\Rightarrow 6mg = 2g$$ Given: $$ T_{\max} - T_{\min} = 2g$$

$$\Rightarrow m = \dfrac{2}{6} = \dfrac{1}{3} kg$$

Hence $$C$$ is the correct option.
Question 8
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If a particle of mass $$M$$ is tied to a light inextensible string fixed at point $$P$$ and particle is projected at A with velocity $$V_A\, =\, \sqrt{4 gL}$$ as shown. Find tension in the string at $$B$$. (Assume particle is projected in the vertical plane.)

A
$$Mg$$

B
$$3Mg$$

C
$$2Mg$$

D
$$7Mg$$

Solution

Taking point A as reference where gravitational potential energy is zero,

Total energy of particle at point A is, $$E_A = \dfrac{1}{2}MV_A^2 = 2MgL$$

Let velocity at point B as $$V_B$$.
So, Total energy at point B is, $$E_B = mgL + \dfrac{1}{2}MV_B^2$$
By conservation of energy, $$E_A=E_B$$
$$\implies \dfrac{1}{2}MV_B^2 = mgL\implies \dfrac{MV_B^2}{L}=2Mg$$

So centripetal force point B is $$2Mg$$
At point B, $$T+Mg=\dfrac{MV_B^2}{L}=2Mg\implies T=Mg$$

So Tension in the string is $$Mg$$, at point B.
Question 9
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A spherical ball A of mass $$4\ kg$$, moving along a straight line strikes another spherical ball B of mass $$1\ kg$$ at rest. After the collision, A and B move with velocities $$v_{1}\ ms^{-1}$$ and $$v_{2}\ ms^{-1}$$ respectively making angles of $$30^{\circ}$$ and $$60^{\circ}$$ with respect to the original direction of motion of ball A. The ratio $$\dfrac {v_{1}}{v_{2}}$$ will be:

A
$$\dfrac {\sqrt {3}}{4}$$

B
$$\dfrac {4}{\sqrt {3}}$$

C
$$\dfrac {1}{\sqrt {3}}$$

D
$$\sqrt {3}$$

Solution

Since there is no momentum in the line perpendicular to the direction of motion of ball A before collision, after collision also there would be no momentum along that line. So by conservation of momentum along that line we have:
$$0=4{ v }_{ 1 }\sin { 30° } -{ v }_{ 2 }\sin { 60° } \Rightarrow \dfrac { { v }_{ 1 } }{ { v }_{ 2 } } =\dfrac { \sqrt { 3 } }{ 4 } $$
So, option A is the correct answer.
Question 10
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A body crosses the topmost point of a vertical circle with critical speed. What will be its centripetal acceleration when the string is horizontal. :-

A
$$\;g$$

B
$$\;2g$$

C
$$\;3g$$

D
$$\;6g$$

Solution

The critical speed of the body at the topmost point is equal to $$\sqrt{gR}$$ i.e $$v = \sqrt{gR}$$
Let the velocity of the body when the string is horizontal be $$v_2$$
Using work-energy theorem : $$W = \Delta K.E$$
$$\therefore$$ $$mgR = \dfrac{1}{2} m v_2^2 - \dfrac{1}{2}mv^2$$
OR $$mgR = \dfrac{1}{2} m v_2^2 - \dfrac{1}{2}m(gR)$$ $$\implies v_2^2 = 3gR$$
Centripetal acceleration $$a_r = \dfrac{v_2^2}{R} = 3g$$