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

Angular velocity Quiz 2

physics

The linear velocity of a given wheelchair is $0.2{m}/{s}\,$. Its angular velocity is,<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/fae2b1ca-f54a-4df3-bb8c-b08f815a152c5948478543271507834.png"/>

 The angular velocity can be found by dividing the linear velocity with the radius of the wheel i.e, $\omega =\dfrac{v}{r}$. The units of the parameters should be taken care of.  

 $\text{The angular velocity of the wheel can be found by the equation,}\\  \omega =\dfrac{v}{r}\\\text{Here, we were given a radius of the wheel and linear velocity.}\\  \text{By substituting this in the given equation, angular velocity will be,}\\  \omega =\dfrac{v}{r} \\  \Rightarrow \omega =\dfrac{0.2}{1}\\  \therefore \omega=0.2\,rad/sec$  

 The kid’s fly wind will rotate 15 times per sec having a radius of 6 cm. Its angular velocity in terms of the rad per sec will be,<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/ba413cc5-4f65-4f95-ac4a-857b6df0419f8719405028476543278.png"/> 

 The radius of wind mill is given to create confusion. The terms rotation per second refers to the frequency. Angular velocity of a body is given by multiplying its frequency with $2\pi $ i.e, $\omega =2\pi f$. 

  $\text{It is given that the toy is rotating 15 times in a second.}\\  \text{So, its frequency is 15 Hz.}\\  \text{Now, the angular velocity of this can be found by using the formula,}\\ \omega =2\pi f \\ \Rightarrow \omega =2\pi \times 15\\ \therefore\omega =30\pi $

A girl is twirling a hula hoop ring of 5 m diameter. If she twirls the ring at 20 revolutions per minute then how far does the ring would fall in 60 minutes?<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/1cf741e5-3363-4e62-b8cf-71743c185cee7602802399530042977.png"/> 

The distance covered by an object/body after completing the revolutions equals the circumference covered by it. Then, the number of revolutions made in 60 minutes multiplied with the circumference will give the distance covered.

 $ \text{Frequency of the revolution is given as 20 rpm.}\\  \text{ That means 20 revolutions are made in a minute.}\\  \text{Then, number of revolutions made in 60 minutes will be,}\\  n=20\times 60=1200\text{ revolutions}\\  \text{Distance covered during a revolution is given by the circumference of the hula hoop ring.}\\  x=2\pi r \\  \Rightarrow x=2\pi \left( \dfrac{d}{2} \right)=5\pi \text{ m} \\\text{Then, in 60 minutes, the ring will be covering a distance of,}\\ {{x}_{total}}=n\times x \\\therefore {{x}_{total}}=1200\times 5\pi =6000\pi \text{ m} $ 

 A delivery boy is riding a scooter having 0.35 m radius tires on his way to deliver at a speed of $3.5{m}/{s}\,$.  A chewing gum has got stuck on the front tire. What is the speed of the blue dot due to chewing gum when it is 0.35 m above the road?<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/a7aa42d1-fa25-4694-aa54-c2774fc199f42547686032733007681.png"/> 

 In this case, the chewing gum will be experiencing both rotational and translational motion. So, its velocity will be given by the resultant of both the vectors. 

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/2c740513-cd08-47b7-9e2b-10bcbcf6c0947706102348109345050.png"/>$ \\ \text{The transitional velocity will be the same as of the velocity of the scooter. That is,}\,3.5{m}/{s}.\\  \text{Now, the angular velocity can be found using the formula,}\\ \omega =\dfrac{v}{r} \\  \Rightarrow\omega =\dfrac{3.5}{0.35}=10\text{ }rad/s \\  \text{Now, the net velocity experienced by the chewing gum will be,}\\  {{v}_{net}}=\sqrt{{{\left( {{v}_{t}} \right)}^{2}}+{{\left( {{v}_{r}} \right)}^{2}}+2{{v}_{t}}{{v}_{r}}\cos \theta } \\  \Rightarrow {{v}_{net}}=\sqrt{{{\left( 3.5 \right)}^{2}}+{{\left( 10 \right)}^{2}}+2\times 3.5\times 10\times \cos \left( 90{}^\circ  \right)} \\  \Rightarrow {{v}_{net}}=\sqrt{{{\left( 3.5 \right)}^{2}}+{{\left( 10 \right)}^{2}}} \\  \Rightarrow {{v}_{net}}=\sqrt{12.25+100}=\sqrt{122.25} \\  \therefore {{v}_{net}}=10.5948\sim 10.6\text{ }m/s $

A dismantled handle of the cycle is mounted on a nearly frictionless axle through its centre.The handle consists of a rod with length 35 cm, and two handgrips of mass 0.25 g. The two horizontal forces of equal magnitude are applied, one facing left applied at a handgrip and the other facing right applied at the centre. At this instant $55{rad}/{s}\,$ will be the angular velocity of the handle and is at an angle $30{}^\circ $ clockwise from the horizontal. What is the magnitude of each force, if in case, in the next 0.3 s, the angular momentum at the centre increases by a value of $1.5\times {{10}^{-4}}\,{kg\,{{m}^{2}}}/{s}\,$?<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/59f3c70e-5220-4ef4-b525-58c8339473ee784064737206063390.png"/>

 The rotational equivalent of the linear force is the torque, so, using the formula relating the torque and equating them, we can find the applied force. We can find torque by taking the cross product between applied force and distance from the point of rotation. Also, torque is defined as the rate of change of angular momentum. 

 &nbsp;$\text{First of all, torque can be found using two formulas.}\\  \text{It is defined as the rate of change of angular momentum. That is,}\\  \tau =\dfrac{dL}{dt}\\  \text{Also, if we take the cross product between the distance from the axis of rotation and the applied force, it will give torque.}\\  \overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}\\ \Rightarrow\overrightarrow{\tau }=r.F.\sin \theta\\  \text{Now, by equating these two relations, we can find the magnitude of applied force.}\\  \dfrac{dL}{dt}=r.F.\sin \theta \\ \Rightarrow \dfrac{dL}{dt}=\left( \dfrac{d}{2} \right).F.\sin \theta  \\  \Rightarrow \dfrac{1.5\times {{10}^{-4}}{kg\,{{m}^{2}}}/{s}\,}{0.3}=\dfrac{0.35}{2}.F.\sin \left( 30{}^\circ  \right) \\  \Rightarrow F=\dfrac{1.5\times {{10}^{-4}}\times 2}{0.3\times 0.35\times \sin \left( 30{}^\circ  \right)} \\  \Rightarrow F=\dfrac{1.5\times {{10}^{-4}}\times 2}{0.3\times 0.35\times \dfrac{1}{2}}\\  \Rightarrow F=\dfrac{6\times {{10}^{-4}}}{3\times 3.5\times {{10}^{-2}}} \\  \Rightarrow F=\dfrac{4}{7}\times {{10}^{-2}}=5.71\times {{10}^{-3}}N \\  \therefore F=5.71\text{ mN} $