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InterviewSolution
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A vehicle of mass `m` starts moving along a horizontal circle of radius R such that its speed varies with distance s covered by the vehicle as `c=Ksqrts`, where K is a constant. Calculate: a. Tangential and normal force on vehicle as fucntion of s. b. Distance s in terms of time t. c. Work done by the resultant force in first t seconds after the beginning of motion. |
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Answer» Correct Answer - a. `(mK^2s)/(R)`; `1/2mK^2`; b. `1/4K^2t^2`; c. `1/8mK^4t^2` Since, the vehicle is moving along a circle, therefore, it is necessarily subjected to a centripetal acceleration. The acceleration `a_n=(v^2)/(R)=(K^2s)/(R)` The normal force `F_n=m.a_n=(mK^2s)/(R)` Since, speed v of particle is increasing, therefore, it is necessary subjected to a tangerial acceleration also, The acceleration `a_t=(d)/(dt)*v=(K)/(2sqrts)*(ds)/(dt)=(K)/(2sqrts)v=(K^2)/(2)` Tangential forces, `F_t=m.a_t=1/2mK^2` Since, s is the distance moved by the particle, therefore, speed v is `ds//dt`. Hence, `(ds)/(dt)=Ksqrts` `(ds)/(sqrts)=Kdt` Integrating the above equation, `underset0oversets int(ds)/(sqrts)=Kint_0^tdt` `2sqrts=Kt` or `s=1/4K^2t^2` `:. Speed v=Ksqrts=1/2K^2t` Work done by resultant force is used to increase kinetic energy of the particle. Kinetic energy at time t, `E=1/2mv^2=1/8mK^4t^2` Initial kinetic energy, `E_0=0`. Increase in kinetic energy `=E-E_0=1/8mK^4t^2` |
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