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A cord of negligible mass is wound round the rim of a fly wheel of mass 20 kg and radius 20 cm. A steady pull of 25 N is applied on the cord as shown in Fig. 7.35. The flywheel is mounted on a horizontal axle with frictionless bearings. (a) Compute the angular acceleration of the wheel. (b) Find the work done by the pull, when 2m of the cord is unwound. (c) Find also the kinetic energy of the wheel at this point. Assume that the wheel starts from rest. (d) Compare answers to parts (b) and (c).

Answer» <html><body><p></p>Solution :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/KPK_AIO_PHY_XI_P1_C07_E02_015_S01.png" width="80%"/> <br/> (a) Torque `<a href="https://interviewquestions.tuteehub.com/tag/tau-663951" style="font-weight:bold;" target="_blank" title="Click to know more about TAU">TAU</a>=RFsintheta` <br/> `=0.2xx25xxsin90^(@)` <br/> `tau=5.0Nm[because sin90^(@)=<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>]` <br/> Moment of inertia `I=(MR^(2))/(2)` <br/> `=(<a href="https://interviewquestions.tuteehub.com/tag/20xx-292987" style="font-weight:bold;" target="_blank" title="Click to know more about 20XX">20XX</a>(0.2)^(2))/(2)` <br/> `therefore I=0.4kgm^(2)` <br/> Now `Ialpha=tau` <br/> `therefore alpha=(tau)/(I)=(5)/(0.4)` <br/> `therefore alpha=12.5"rad/s"^(2)""....(1)` <br/> (b) Work <a href="https://interviewquestions.tuteehub.com/tag/done-2591742" style="font-weight:bold;" target="_blank" title="Click to know more about DONE">DONE</a> by the pull unwinding 2m of the cord <br/> `W=Fl` <br/> `=25xx2` <br/> `therefore W=50J` <br/> (c ) Let `omega` be the final <a href="https://interviewquestions.tuteehub.com/tag/angular-11524" style="font-weight:bold;" target="_blank" title="Click to know more about ANGULAR">ANGULAR</a> velocity <br/> The kinetic energy gained `=(1)/(2)Iomega^(2)...(2)` <br/> If wheel gains angular velocity `omega` after starting from rest, then <br/> Now `omega^(2)=omega_(0)^(2)+2alphatheta` <br/> `therefore omega^(2)=2alphatheta""[because omega_(0)=0]...(3)` <br/> Angular displacement <br/> `theta=("length of unwound string")/("radius of wheel")` <br/> `=(2)/(0.2)` <br/> `therefore theta=10` rad `""....(4)` <br/> From eqn. (1), (3) and (4) <br/> `omega^(2)=2alphtheta` <br/> `=2xx12.5xx10` <br/> `therefore omega^(2)=250 ("rad s"^(-1))^(2)` <br/> `therefore` kinetic energy gained <br/> `K.=(1)/(2)Iomega^(2)` <br/> `=(1)/(2)xx0.4xx250` <br/> `therefore K=50J""...(5)` <br/> (d) Equation (2) and (5) are equal, means kinetic energy gained by the wheel = work done by the force. <br/> Here, there is no loss of energy due to friction.</body></html>


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