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This is second of our many quizzes on Rotational motion. Click on this to access more such competitive quizzes on Mechanics.

Rotational motion Quiz 2

physics

A disc of mass $M$ and radius $R$ is rotating with angular  ${{\omega }_{o}}$ about an axis perpendicular passing through its centre. Three equal masses $'m'$ are placed on the disc. One mass is kept at centre, while the other two at opposite extremes of disc. The new angular speed due to this will be-  

There is no external torque on the system, so the angular momentum remains conserved. 

  As there is no external torque on the system, the angular momentum remains conserved $ \\{L}_{\operatorname{initial}}={{L}_{final}} \\ \Rightarrow {{I}_{i}}{{\omega }_{i}}={{I}_{f}}{{\omega }_{f}} \\ $ As ${{I}_{i}}$ is the moment of inertia of disc only, then $ {{I}_{i}}=\dfrac{M{{R}^{2}}}{2}$ $($about axis passing through its centre perpendicularly$)$ $ \\ \Rightarrow {{\omega }_{i}}={{\omega }_{o}}\text{(given)} \\ \Rightarrow {{\text{I}}_{f}}={{I}_{disc}}+{{I}_{mass}}+{{I}_{mass}} \\ \Rightarrow {{I}_{f}}={{I}_{disc}}+\left( m{{R}^{2}} \right)+\left( m{{R}^{2}} \right) \\ \Rightarrow {{I}_{f}}=\dfrac{M{{R}^{2}}}{2}+2m{{R}^{2}} \\ \Rightarrow {{I}_{f}}={{R}^{2}}\left( \dfrac{M}{2}+2m \right) \\Thus \\ \left( \dfrac{M{{R}^{2}}}{2} \right){{\omega }_{o}}={{R}^{2}}\left( \dfrac{M}{2}+2m \right){{\omega }_{f}} \\ \Rightarrow{{\omega }_{f}}=\dfrac{M{{\omega }_{o}}}{2\left( \dfrac{M}{2}+2m \right)} \\ \therefore{{\omega }_{f}}=\dfrac{M{{\omega }_{o}}}{\left( M+4m \right)}$

 A rod is hung to the ceiling freely so it's able to rotate in a plane perpendicular to an axis passing through one of its ends. A bullet of mass $'m'$ with velocity ${{V}_{o}}$ strikes the rod elastically on its free end. If the mass of the rod is $'M'$ and it’s length is $'\ell '$, then the angular velocity of the rod will be<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/84440f64-a71b-4710-84be-a125ea9862625644615152428048368.png"/> 

Angular momentum is conserved, so calculate the final angular momentum using the formula, $(\dfrac{M{{\ell }^{2}}}{3}+m{{\ell }^{2}})\omega$

  As the are no external torque acting on the system, the angular momentum remains conserved $ \\ $ Initial angular momentum of rod $= 0$ $ \\ $  Initial angular momentum of bullet just before it strikes the rod $ =m\times {{V}_{o}}\times \ell $ $ \\ $ If both the bullet and rod get final angular velocity $\omega$ then, $ \\ $ Final moment of inertia of the system $ = {{\text{I}}_{rod}}+{{I}_{bullet}}=\dfrac{M{{\ell }^{2}}}{3}+m{{\ell }^{2}} \\ $ Final angular momentum $ = \left( \dfrac{M{{\ell }^{2}}}{3}+m{{\ell }^{2}} \right)\omega  \\ \text{As }{{\text{L}}_{i}}={{L}_{f}} \\m\times {{V}_{o}}\times \ell =\left( \dfrac{M}{3}+m \right){{\ell }^{2}}\omega  \\ \Rightarrow\omega =\dfrac{m{{V}_{o}}}{\left( \dfrac{M}{3}+m \right)\ell } \\ \therefore\omega =\dfrac{3m{{V}_{o}}}{\left( M+3m \right)\ell }$

Pulley having mass $''m''$ and radius $''R''$ is free to rotate about the fixed point as shown in figure.<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/b6170823-1510-4128-99b3-6bb2b3ec3ca67111448648564833776.png"/>$\\$ The acceleration of $''2m''$ block is

 Draw free body diagram and calculate the value of ${{T}_{1}}$ and ${{T}_{2}}$ and use $\left( {{T}_{1}}-{{T}_{2}} \right)R=I\alpha $. 

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/807670b3-6cb4-40c0-a2ac-8752c68ac3151715728830266417891.png"/> $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/89d90934-b79e-4617-ac7b-51aa5eaaf9ec9184764231786899276.png"/>$ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/246827f2-f3ec-4632-bc70-609ceb7b2f588099716437324392949.png"/> $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/a11e86ed-4a98-4930-bc83-fe3c1b1fdd793195108453161805520.png"/>$ \\ \text{From figure,} \\2mg-{{T}_{1}}=2ma..........\left( i \right) \\ \Rightarrow\left( {{T}_{1}}-{{T}_{2}} \right)R=\dfrac{m{{R}^{2}}}{2}\alpha  \\ \Rightarrow\left( {{T}_{1}}-{{T}_{2}} \right)=\dfrac{ma}{2}..........\left( ii \right) \\ \left( \because a=\alpha R \right) \\ \Rightarrow{{T}_{2}}-mg\sin \theta =ma..........\left( iii \right) \\ \text{Adding }\left( \text{i} \right)\text{ and }\left( \text{iii} \right)\text{,} \\2mg-\left( {{T}_{1}}-{{T}_{2}} \right)-mg\sin \theta =3ma \\ \text{Putting }{{T}_{1}}-{{T}_{2}}=\dfrac{ma}{2} \\ \Rightarrow mg\left( 2-\sin \theta  \right)=\dfrac{7}{2}ma \\ \therefore a=\dfrac{2}{7}g\left( 2-\sin \theta  \right)$

In a figure given below, a rod of mass $''M''$ and length $''L''$ is supported by threads with tension $''{{T}_{1}}''$ and $''{{T}_{2}}''$. A block of mass $\left( \dfrac{M}{4} \right)$ is suspended by a thread of tension $''T''$. If the system is in equilibrium then find tension $'{{T}_{1}}'$. $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/ebb1bfb6-9ee8-4999-ba8b-ed4d6c856da41374173042315328621.png"/> 

Draw the free body diagram and balance the force in vertical direction to calculate the tension values. 

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/4bee7e86-60d1-4a90-94d8-1e2d693188f58901594100041748469.png"/> $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/9ed49058-4e19-4ee0-bb42-88b1ac7ea2143424722496217169163.png"/> $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/ba54e769-5f29-4dd0-a9b5-650caa939f7f6145904290562559037.png"/><br>$ \\ $ From figure $(1)$ $ \\{{T}_{1}}+{{T}_{2}}=Mg+T \\ $ From figure $(2)$ $T=\dfrac{Mg}{4} \\ \Rightarrow {{T}_{2}}=\dfrac{5}{4}Mg-{{T}_{1}} \\ $  Since system is in equilibrium, balanced torque around C.O.M is $ \\ {{\tau }_{{{T}_{1}}}}+{{\tau }_{T}}={{\tau }_{{{T}_{2}}}} \\ \Rightarrow\dfrac{L}{2}\times {{T}_{1}}+\dfrac{L}{2}\times T=\dfrac{L}{2}\times {{T}_{2}} \\ \Rightarrow {{T}_{1}}+T=\dfrac{{{T}_{2}}}{1} \\ \Rightarrow {{T}_{1}}+\dfrac{M}{4}g=\dfrac{5}{4}Mg-{{T}_{1}} \\ \Rightarrow 2{{T}_{1}}=Mg \\ \therefore {{T}_{1}}=\dfrac{Mg}{2}$

 In figure there are three identical discs of mass $''M''$ and radius $''R''$ kept as shown. The moment of inertia about axis $X-X'$ is $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/7b08fc6a-5f42-48b4-9256-41f6dc7312618550298605309354281.png"/> 

 Calculate moment of inertia about diagonal axis of geometry then about $yy'$ axis.Moment of inertia of the diagonal axis is $\dfrac{M{{R}^{2}}}{4}$. 

 Moment of Inertia of a solid disc axis passing diametrically to its plane is $ =\dfrac{M{{R}^{2}}}{4}$ $ \\ $ Using parallel axis theorem, $ \\ $ Total moment of Inertia about $XX’$. $ \\ {{I}_{XX'}}=\left( {{I}_{A}}+M{{R}^{2}} \right)+\left( {{I}_{B}}+M{{\left( 3R \right)}^{2}} \right)+\left( {{I}_{C}}+M{{\left( 2R \right)}^{2}} \right) \\ \Rightarrow{{I}_{XX'}}=\dfrac{M{{R}^{2}}}{4}+\dfrac{M{{R}^{2}}}{4}+\dfrac{M{{R}^{2}}}{4}+14M{{R}^{2}} \\ \therefore {{I}_{XX'}}=\dfrac{59}{4}M{{R}^{2}}$