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This is one of our many quizzes on Angular acceleration. Click on this to access more such competitive quizzes on Circular Motion.

Angular acceleration Quiz 2

physics

Velocity of a reaction wheel changes from $\pi \text{ rad }{{\text{s}}^{-1}}$ to $2\pi \text{ rad }{{\text{s}}^{-1}}$. If the torque and work done on the reaction wheel are $0.0628\text{ N m}$ and $0.3\text{ J}$ respectively, determine the angular acceleration linked with the reaction wheel. 

 Determining moment of inertia linked with the reaction wheel will help to arrive at the correct answer. 

  $\text{From work done on the reaction wheel, moment of inertia can be derived as:}\\W=\dfrac{1}{2}I({{\omega }_{2}}^{2}-{{\omega }_{1}}^{2}) \\\Rightarrow 0.3=\dfrac{1}{2}I({{\omega }_{2}}^{2}-{{\omega }_{1}}^{2}) \\\Rightarrow I=\dfrac{0.6}{({{\omega }_{2}}^{2}-{{\omega }_{1}}^{2})} \\\Rightarrow I=\dfrac{0.6}{(4{{\pi }^{2}}-{{\pi }^{2}})}\text{ } \\\Rightarrow I=\dfrac{0.6}{3{{\pi }^{2}}} \\\Rightarrow I=0.02\text{ kg }{{\text{m}}^{2}}\text{}\to (1) \\\text{Now, torque on the reaction wheel is given by}\\\tau =I\alpha \text{}\to \text{(2)}\\\text{From (1) and (2), angular acceleration linked with the reaction wheel is:}\\\alpha =\dfrac{\tau }{I} \\ \Rightarrow \alpha =\dfrac{0.0628}{0.02} \\\therefore \alpha =3.14\text{ rad }{{\text{s}}^{-2}}$

A tyre rotates from $27\text{ rpm}$ to $'x'\text{ rpm}$ in one quarter of a minute. If the ratio of torque and rotational inertia linked with the rotating tyre is $3:4$, determine the value of $'x'$. 

 Recall and combine the expressions for angular acceleration to arrive at the required frequency of tyre. 

  $\text{Ratio of torque and rotational inertia is nothing but angular acceleration.}\\\tau =I\alpha  \\\Rightarrow \dfrac{\tau }{I}=\dfrac{3}{4} \\\Rightarrow \alpha =0.75\text{ rad }{{\text{s}}^{-2}} \\\text{Initial angular velocity of the tyre is given by}\\{{\omega }_{i}}=2\pi {{n}_{i}} \\\Rightarrow {{\omega }_{i}}=2\pi \times \dfrac{30}{60} \\\Rightarrow {{\omega }_{i}}=\pi \text{ rad }{{\text{s}}^{-1}} \\\text{Final velocity if the tyre is given by}\\{{\omega }_{f}}=2\pi {{n}_{f}} \\\Rightarrow {{\omega }_{f}}=2\pi \times \dfrac{x}{60} \\\Rightarrow {{\omega }_{f}}=\dfrac{\pi x}{30}\text{ rad }{{\text{s}}^{-1}} \\\text{Angular acceleration is given by}\\\alpha =\dfrac{{{\omega }_{f}}-{{\omega }_{i}}}{t} \\\Rightarrow \alpha =\dfrac{\left( \dfrac{\pi x}{30}-\pi  \right)}{\left( \dfrac{1}{4}\times 60 \right)} \\\Rightarrow 0.75=\dfrac{\left( \dfrac{\pi x}{30}-\pi  \right)}{15} \\\Rightarrow \dfrac{\pi x}{30}-\pi =11.25 \\\Rightarrow x=\dfrac{11.25\times 30}{\pi }+30 \\\therefore x=137.48\approx 137\text{ rpm}$

Suppose that a body of mass $3.2\text{ g}$ is spinning around a particular point with a torque of $7.2\times {{10}^{-9}}\text{ Nm}$. The gyradius of the spinning body is given to be $2.1\text{ mm}$.Determine the body’s angular acceleration. 

Gyradius or radius of gyration of a spinning body is directly proportional to the square of moment of inertia linked with the body.  

 $\text{Gyradius or radius of gyration of the spinning body is given to be:}\\K=2.1\text{ mm}=2.1\times {{10}^{-3}}\text{ m}\\\text{Mass of the body is given to be:}\\M=3.2\text{ mg}=3.2\times {{10}^{-3}}\text{ kg}\\\text{Moment of inertia of the body is given by}\\I=M{{K}^{2}} \\\Rightarrow I=3.2\times {{10}^{-3}}\times {{(2.1\times {{10}^{-3}})}^{2}} \\\Rightarrow I=14.112\times {{10}^{-9}}\text{ kg }{{\text{m}}^{2}} \\\text{Moment of inertia is related to torque by}\\I=\tau \alpha\\\text{Therefore, angular acceleration can be written as}\\\alpha =\dfrac{I}{\tau } \\\Rightarrow \alpha =\dfrac{14.112\times {{10}^{-9}}}{7.2\times {{10}^{-9}}} \\\therefore \alpha =1.96\text{ rad }{{\text{s}}^{-2}}$

The ratio of kinetic energy to the angular momentum on a discus is $2:3$ at an instant given by $t=28\text{ sec}$. If the same at an instant given by $t=34\text{ sec}$ is $1:4$, evaluate the angular acceleration acquired by the discus during this time interval.  

Determine the expression for the ratio of given quantities to evaluate the angular acceleration during the given time interval. 

 $\text{Kinetic energy of discus at a particular instant is given by}\\KE=\dfrac{1}{2}I{{\omega }^{2}}\\\text{Angular momentum of the discus is given by}\\L=I\omega\\\text{Clearly, ratio of kinetic energy and angular momentum is given by}\\\dfrac{KE}{L}=\dfrac{\left( \dfrac{1}{2}I{{\omega }^{2}} \right)}{I\omega } \\\Rightarrow \dfrac{KE}{L}=\dfrac{\omega }{2} \\\text{Involving the data given in the question at t=28 sec},\\\dfrac{{{\omega }_{i}}}{2}=\dfrac{2}{3} \\\Rightarrow {{\omega }_{i}}=\dfrac{4}{3}\text{ rad }{{\text{s}}^{-1}} \\\text{Similarly, involving the data given in the question at}\,t=34\text{ sec},\\\dfrac{{{\omega }_{f}}}{2}=\dfrac{1}{4} \\\Rightarrow {{\omega }_{f}}=\dfrac{1}{2}\text{ rad }{{\text{s}}^{-1}} \\\text{Clearly, angular acceleration acquired by the discus during the given time interval can be derived as:}\\\alpha =\dfrac{{{\omega }_{f}}-{{\omega }_{i}}}{{{t}_{2}}-{{t}_{1}}} \\\Rightarrow \alpha =\dfrac{\left( \dfrac{4}{3}-\dfrac{1}{2} \right)}{34-28} \\\therefore \alpha =\dfrac{5}{36}\text{ rad }{{\text{s}}^{-2}}$

A frisbee is rotating with an angular velocity of $\text{8 rad }{{\text{s}}^{-1}}$. Weight and diameter of the frisbee are $\text{3250 g}$ and $\text{30 cm}$. Suppose two bar magnets of mass $25\text{ g}$ each are gently kept at the ends of one of the diameters of the frisbee with the help of a drone, what would be the magnitude of angular acceleration linked with the frisbee in a second?  

 Involve the concept of conservation of angular momentum to determine the final angular velocity, which can further be used to find the magnitude of the required quantity. 

 $\text{Initial angular velocity of the frisbee is given as}\\{{\omega }_{i}}=8\text{ rad }{{\text{s}}^{-1}}\\\text{Mass of frisbee from its weight can be derived as}\\W=Mg \\\Rightarrow M=\dfrac{W}{g} \\\Rightarrow M=\dfrac{3250}{10} \\\Rightarrow M=325\times {{10}^{-3}}kg \\\text{Radius of the frisbee from its diameter can be derived as}\\D=2R \\\Rightarrow R=\dfrac{D}{2} \\\Rightarrow R=\dfrac{30}{2} \\\Rightarrow R=15\times {{10}^{-2}}\text{ m} \\\text{Now, it is said that two bar magnets of mass}\,m=25\times {{10}^{-3}}\text{ kg}\,\text{are gently kept placed at the ends of a diameter of the frisbee.}\\\text{Here, conservation of angular momentum applies as follows:}\\{{I}_{i}}{{\omega }_{i}}={{I}_{f}}{{\omega }_{f}} \\\Rightarrow \dfrac{M{{R}^{2}}}{2}\times {{\omega }_{i}}=\left( \dfrac{M{{R}^{2}}}{2}+2m{{R}^{2}} \right)\times {{\omega }_{f}} \\\Rightarrow \dfrac{M{{R}^{2}}}{2}\times {{\omega }_{i}}=\left( \dfrac{M{{R}^{2}}+4m{{R}^{2}}}{2} \right)\times {{\omega }_{f}} \\\Rightarrow {{\omega }_{f}}=\dfrac{M{{\omega }_{i}}}{(M+4m)} \\\Rightarrow {{\omega }_{f}}=\dfrac{325\times {{10}^{-3}}\times 8}{(325\times {{10}^{-3}}+(4\times 25\times {{10}^{-3}}))} \\\Rightarrow {{\omega }_{f}}=\dfrac{2600}{425} \\\Rightarrow {{\omega }_{f}}=6.11\text{ rad }{{\text{s}}^{-1}} \\\text{Now, magnitude of angular acceleration linked with the frisbee in a second can be derived as:}\\\alpha =\dfrac{{{\omega }_{f}}-{{\omega }_{i}}}{t} \\\Rightarrow \alpha =\dfrac{6.11-8}{1} \\\therefore \left| \alpha  \right|=1.89\text{ rad }{{\text{s}}^{-2}}$ 