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Derive the expression for final speed of a particle moving in an inclined plane. |
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Answer» Solution :(i) Let us assume a round object of mass m and radius R is rolling down an inclined plane without slipping as shown in Figure. There are two forces acting on the object along the inclined plane. (ii) One is the component of gravitational force `(mgsintheta)` and the other is the static frictional force (f). The other component of gravitation force `(mgcostheta)` is cancelled by the normal force (N) exerted by the plane. As the motion is happening along the incline, we shall write the equation for motion from the thefree body diagram (FBP) of the object.  (iii) For translational motion, `mg sintheta` is the supporting force and f is the OPPOSING force, `""mgsintheta-f=ma` For rotational motion, let us take the TORQUE with respect to the center of the object. Then `mg sintheta` cannot cause torque as it passes through it but the frictionalforce f can set torqueof Rf. `""Rf=Ialpha` (iv) By using the relation, `a=ralpha`, and momentof inertia `I=mK^(2)`, we get, `Rf=mK^(2)a/R, f=ma(K^(2)/R^(2))` Now equation becomes, `mgsintheta-ma(K^(2)/R^(2))=ma` `mgsintheta=ma+ma(K^(2)/R^(2))` `a(1+K^(2)/R^(2))=gsintheta` After rewriting it for acceleration, we get, `a=(gsintheta)/((1+K^(2)/R^(2)))` (v) We can ALSO find the expression for final velocity of the rolling object by using third equation of motion for the inclined plane `v^(2)=u^(2)+2as`. If the body starts rolling from rest, u=0. When h is the vertical height of the incline, the length of the incline s is, `s=h/sintheta` `v^(2)=2(gsintheta)/((1+K^(2)/R^(2)))(h/sintheta)=(2gh)/((1+K^(2)/R^(2)))` By taking square root, `v=sqrt((2gh)/((1+K^(2)/R^(2))))` (vi) The time taken for rolling down the incline could also be written from first equation of motion as, v=u+at. For the object which starts rolling from rest, u=0. Then, `t=v/a` `t=(sqrt((2gh)/((1+K^(2)/R^(2)))))(((1+K^(2)/R^(2)))/(gsintheta))` `t=sqrt((2h(1_+K^(2)/R^(2)))/(gsin^(2)theta))` (vii) The equation suggests that for a given incline, the object with the LEAST value of radius of gyration K will reach the bottom of the incline first. |
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