Free Study Material

NCERT Solutions

NCERT Books

Revision Notes

Practice Problems

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

Torque Quiz 2

physics

The initial value of angular velocity is $2\text{ units}$. If the magnitude of torque is $1\text{ unit}$ and the change in time is $4\text{ sec}$ then determine the final value of angular velocity.What would happen if the torque magnitude was zero?

 The final angular velocities for both the cases can easily be determined from the mathematical expression relating torque and the change in angular velocity. 

  $\text{Given}\,{{\omega }_{i}}=2\text{ units} ; \tau =1\text{ unit}\,\text{and}\,t=4\text{ sec}\\\text{We know that torque is given by}\\\tau =\dfrac{{{\omega }_{f}}-{{\omega }_{i}}}{t} \\\Rightarrow {{\omega }_{f}}-{{\omega }_{i}}=\tau t \\\Rightarrow {{\omega }_{f}}={{\omega }_{i}}+\tau t \\\Rightarrow {{\omega }_{f}}=2+1\times 4 \\\Rightarrow {{\omega }_{f}}=6\text{ units} \\\text{Moving on to the next case,<br>if}\,\tau =0,\\{{\omega }_{f}}={{\omega }_{i}}+\tau t \\\Rightarrow {{\omega }_{f}}=2+0\times 4 \\\Rightarrow {{\omega }_{f}}=2\text{ units}$

A flywheel has an angular velocity given by $3{{t}^{2}}+2t$. Determine the torque acting on the flywheel at $t=10\text{ sec}$. Take the moment of inertia of the flywheel to be $20\text{ units}$. 

 Torque is directly proportional to moment of inertia as well as angular acceleration. 

  $\text{Given, angular velocity}\,\omega =3{{t}^{2}}+2t\\\text{We know that angular acceleration is given by}\\\alpha =\dfrac{d\omega }{dt} \\\Rightarrow \alpha =6t+2 \\\text{At}\,t=10\text{ sec}\\\alpha =6\times 10+2 \\\Rightarrow \alpha =62\text{ units} \\ \text{Now, we also know that torque is given by}\\\tau =I\alpha  \\\Rightarrow \tau =20\times 62=1240\text{ units}$

The work done on a rotating water wheel is $10{{\pi }^{2}}\text{ J}$. If the torque and angular velocity acting on the water wheel are $12\text{ N m}$ and ${{\pi }^{2}}\text{ rad }{{\text{s}}^{-1}}$, determine the change in time required for this to happen.

The required time is the ratio of change in angular displacement to the angular velocity. Torque and change in angular displacement are inversely proportional to each other. 

$\text{Given, work done}\,W=10{{\pi }^{2}}\text{ J};\text{torque}\,\tau =12\text{ N m}\,\text{and angular velocity,}\,\omega ={{\pi }^{2}}\text{ rad }{{\text{s}}^{-1}}\\\text{We know that}\\W=\tau d\theta \\\text{Clearly, change in angular displacement is given by}\\d\theta =\dfrac{W}{\tau } \\\Rightarrow d\theta =\dfrac{10{{\pi }^{2}}}{12}\text{ rad} \\\text{Now, we know that angular velocity is given by}\\\omega =\dfrac{d\theta }{dt}\\\text{Clearly, change in time required for the given process to happen on the water wheel is}\\dt=\dfrac{d\theta }{\omega } \\\Rightarrow dt=\dfrac{\left( \dfrac{10{{\pi }^{2}}}{12} \right)}{{{\pi }^{2}}}=\dfrac{10}{12} \\\Rightarrow dt=0.83\text{ s}$

 If the ratio of angular momentum and instantaneous power of a rotating caster is given by $24\text{ W s k}{{\text{g}}^{-1}}\text{ }{{\text{m}}^{-2}}\text{ }$ and if the moment of inertia of the same is given by $4.8\text{ kg }{{\text{m}}^{2}}$, determine the torque of the rotating caster.

 Torque of the rotating caster can easily be determined from the ratio of expressions for angular momentum and instantaneous power. 

 $\text{We know that instantaneous power,}\,P=\tau \omega\,\text{and angular momentum,}\,L=I\omega\\\dfrac{L}{P}=24\text{ W s k}{{\text{g}}^{-1}}\text{ }{{\text{m}}^{-2}}\text{ } \\\Rightarrow \dfrac{I\omega }{\tau \omega }=24 \\\Rightarrow \dfrac{I}{\tau }=24 \\\Rightarrow \dfrac{I}{24}=\tau  \\\Rightarrow \tau =\dfrac{4.8}{24}=0.2\text{ N m}$

 A solid sphere of radius $r$ and mass $M$ is rotated about its diameter in a counter-clockwise direction as shown in the following figure. If the torque acting on the system is $8\text{ Nm}$, deduce an expression for angular acceleration of the solid sphere.<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/bd7b85b5-ecb1-4999-9e55-8bb54ed3902c6547665249127512443.png"/>

 The respective formula for moment of inertia for a solid sphere rotating about its diameter needs to be recalled and substituted in the expression for torque.

 $\text{For a solid sphere rotating about its diameter as shown,} \\ \text{the moment of inertia is given by}\\{{I}_{solid\text{ }sphere}}=\dfrac{2M{{r}^{2}}}{5}\\\text{Now, we know that torque is given by}\\\tau =I\alpha\\\text{Clearly, angular acceleration is given by}\\\alpha =\dfrac{\tau }{I} \\\Rightarrow \alpha =\dfrac{8}{\left( \dfrac{2M{{r}^{2}}}{5} \right)} \\\Rightarrow \alpha =\dfrac{20}{M{{r}^{2}}}\text{ rad }{{\text{s}}^{-2}}$