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This is one of our many quizzes on Newton's second law for rotation. Click on this to access more such competitive quizzes on Laws of Motion.

Newton's second law for rotation Quiz 2

physics

Three balls (A, B, C) are attached with a string on a table whose entire mass is $18\,\text{kg}$ and they are at first very still. Likewise out of three balls, just one ball has an alternate mass. After at some point when the string is opened, at that point, the three balls begin moving and balls whose mass are the equivalents are oppositely moving concerning each other with a speed of $45\,\text{m}{{\text{s}}^{-1}}$. On the off chance that the mass of the three balls after opening the string are in the proportion of $2:2:5$ at that point ascertain the speed of the heavier ball. Expect that the string has no mass. 

 Use the law of conservation of momentum to the momenta to find the velocity of the heavier ball.

  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/65e8d1a2-ac0e-4a4a-8ed7-2531a93148993080047635466090592.png"/> $ \\ $ Let the weight of the three balls $(A, B, C)$ are $ {{\text{m}}_{\text{1}}}\text{, }{{\text{m}}_{\text{2}}}\text{, }{{\text{m}}_{\text{3}}}\\ $ And their corresponding velocities : $ \\ {{\text{v}}_{\text{1}}}\text{,}{{\text{v}}_{\text{2}}}\text{,}{{\text{v}}_{\text{3}}}\\ $ According to the question $ \\ {{m}_{1}}+{{m}_{2}}+{{m}_{3}}=18\text{ kg}\\ $ Also, the ratio is given $ =\text{ }2:2:5\\\Rightarrow 2x+2x+5x=18 \\\text{So }{{m}_{1}}={{m}_{2}}=4\,\text{kg }and\text{ }{{m}_{3}}=10\,\text{kg} \\{{v}_{1}}={{v}_{2}}=45\,\text{m}{{\text{s}}^{\text{-1}}} \\ $ Apply the concept of the law of conservation of momentum, to the momenta along the horizontal direction, $ \\ {{m}_{3}}{{v}_{3}}={{m}_{1}}{{v}_{1}}\cos 45{}^\circ +{{m}_{2}}{{v}_{2}}\cos 45{}^\circ  \\\Rightarrow {{m}_{1}}={{m}_{2}}=4\,\text{kg and }{{m}_{3}}=10\,\text{kg},\text{ }{{v}_{1}}={{v}_{2}}=45\,\text{m}{{\text{s}}^{-1}} \\\Rightarrow 10{{v}_{3}}=4\times 45\times \dfrac{1}{\sqrt{2}}+4\times 45\times \dfrac{1}{\sqrt{2}} \\\Rightarrow {{v}_{3}}=2\times 4\times 45\times \dfrac{1}{\sqrt{2}}\times \dfrac{1}{10}\\\therefore {{v}_{3}}=25.4\,\text{m}{{\text{s}}^{-1}}$

Rahul pass his sponge ball to Alex and at that time the angular velocity of the sponge ball is $\omega $, When Alex pass that sponge ball to Ravi then the size of the ball get reduced without changing its mass and now the radius of the ball becomes $\dfrac{4}{3}$, then what will the new angular velocity of the ball if the external force acting on the ball is constant in both of the cases.  

Apply conservation of angular momentum and with the help of that find the new angular velocity value. 

  Use the concept of conservation of angular momentum: $ \\{{I}_{1}}{{\omega }_{1}}={{I}_{2}}{{\omega }_{2}} \\ $ Moment of inertia for a sphere $ =\dfrac{2}{5}M{{R}^{2}} \\\Rightarrow \dfrac{2}{5}M{{R}_{1}}^{2}\omega =\dfrac{2}{5}M{{R}_{2}}^{2}{{\omega }_{2}} \\\text{Also }{{R}_{2}}=\dfrac{4}{3}{{R}_{1}} \\\Rightarrow \dfrac{2}{5}M{{R}_{1}}^{2}\omega =\dfrac{2}{5}M\dfrac{16}{9}{{R}_{1}}^{2}{{\omega }_{2}} \\\Rightarrow {{\omega }_{2}}=\dfrac{9}{16}\omega  \\\therefore{{\omega }_{2}}=0.5625\,\omega$  

Reena took three long threads and one ball whose mass is $4\,\text{kg}$. She tied that ball with one end of both of the threads and the other ends of both the threads were tied with the third thread then She tries to balance the ball with the help of the third thread so that it must-stops when she hangs the ball, so find Tension in both of the string which was tied with the ball at the time when she able to balance the ball. If the first thread makes an angle $\text{60 }\!\!{}^\circ\!\!\text{ }$ with the third thread and the second thread makes an angle $\text{30 }\!\!{}^\circ\!\!\text{  }\!\!~\!\!\text{ }$ with the third thread when the ball is suspended $\left( g=10\text{ m}{{\text{s}}^{-2}} \right)$. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/a75d4692-d65c-48b1-8cc0-145246aae4c87815886852902643076.png"/> 

First, make the free body diagram and then take vertical and horizontal components where all forces act as zero in the net and then calculate the value of the two tensions. 

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/0ab174d1-7c1a-4457-9c9c-c076fc1ce3d78423952342459634068.png"/> $ \\ $ Let Tension in the first and second thread is ${{T}_{1}}$ and ${{T}_{2}}$ $ \\ $ Since the ball is in equilibrium ,the various forces act on ball must be add to zero, $ \\ $ Taking horizontal components: $ \\{{T}_{1}}\cos \alpha ={{T}_{2}}\cos \beta  \\\Rightarrow {{T}_{2}}={{T}_{1}}\dfrac{\cos \alpha }{\cos \beta }\,\,\to \text{(i)} \\ $ Taking Vertical components : $ \\\Rightarrow {{T}_{1}}\sin \alpha +{{T}_{2}}\sin \beta =mg \\ $ Using eq.$(i)$ $\\ \Rightarrow {{T}_{1}}=\dfrac{mg\cos \beta }{\sin \alpha \cos \beta +\cos \alpha \sin \beta }\\\Rightarrow {{T}_{1}} =\dfrac{mg\cos \beta }{\sin (\alpha +\beta )}\\\Rightarrow {{T}_{1}} =\dfrac{4\times 10\times \cos ({{30}^{0}})}{\sin ({{60}^{0}}+{{30}^{0}})} \\\Rightarrow {{T}_{1}} =20\,\text{N} \\ $ Similarly, $ \\{{T}_{2}}=20\sqrt{3}\,\text{N}$

Ravina and Avi $($both having the same mass$)$ are playing a game ‘kikli $($a game in which two-person hold their hands and start moving in a circular direction with some radius.$)$ If the mass of the Ravina is  $46\,\text{kg}$ and the radius of the circle which is made while playing Kikli is $135\text{ cm}$ then what will be the value of the angular momentum of the Avi.$(\text{ Assume that the velocity is} 36.5\,m {s}^{-1})$. 

Since Ravina and Avi have the same mass then assume the mass of Ravina equals Avi and then calculate angular momentum. 

 Angular momentum $(L)=mvr \\m=46\text{ kg} \\\Rightarrow v=36.5\,\text{m}{{\text{s}}^{\text{-1}}}\text{ } \\\Rightarrow r=135\text{ cm}=1.35\text{ m} \\\text{So put all these values } \\\Rightarrow L=46\times 36.5\times 1.35 \\\therefore L=22666.5\text{ Js}$

While playing in the park when Kashish rides on the Sliding swing her ring come out from her finger and the rings start moving on the sliding swing which is inclined at an angle of ${{60}^{\circ }}$ from the ground If the radius of the ring is $5.63\text{ cm}$, then finding the velocity of the ring on the swing after $3\text{ sec}$. $\text{Take }g=10\text{ m}{{\text{s}}^{-2}}$ 

Apply the concept of acceleration of the body when it is inclined at the plane and then calculate final velocity. 

 Use $ \\a=\dfrac{g\sin \theta }{1+\dfrac{I}{M{{R}^{2}}}} \\ $ Also moment of inertia of the circular ring $ =\dfrac{1}{2}M{{R}^{2}} \\a=\dfrac{10\times \sin ({{60}^{o}})}{1+\dfrac{M{{R}^{2}}}{2M{{R}^{2}}}}=\dfrac{10}{3} \\\Rightarrow v=u+at \\\Rightarrow v=0+(\dfrac{10}{3}\times 3)\\\therefore v=10\,\text{m}{{\text{s}}^{-1}}$