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# A round uniform body of radius R, mass M and moment of inertia I rolls down (without slipping) an inclined plane making an angle θ with the horizontal. Then its acceleration is

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## Step1: Given dataThe radius of the body is R.The mass of the body is M.The moment of inertia of the body is I.The angle of the inclined plane is $\theta$.Step2: Moment of inertia and kinetic energyThe moment of inertia of a body about a given axis in space is the sum of the products of the mass and square of the distance from the axis for each particle comprising the body.The rotational kinetic energy of a body of mass M rotating with an angular velocity $\omega$ is defined by the form, ${E}_{k}=\frac{1}{2}I{\omega }^{2}$, where, I is the moment of inertia. Similarly, the translational kinetic energy of a body of mass with linear velocity $v$ is ${E}_{}=\frac{1}{2}M{v}^{2}$.Step3: Conservation of energyWe know when a body rolls on an inclined plane, then the energy is conserved at every point on the inclined plane. If there is no friction during rolling, $totallossinpotenialenergy=totalgaininkineticenergy.$The potential energy of a body of mass at height is defined by the form, $U=mgh$, where g is the acceleration due to gravity.Step4: Finding the velocityFrom the conservation of energy,$Mgh=\frac{1}{2}M{v}^{2}+\frac{1}{2}I{\omega }^{2}\phantom{\rule{0ex}{0ex}}orMgh=\frac{1}{2}M{v}^{2}+\frac{1}{2}.I{\left(\frac{v}{r}\right)}^{2}\phantom{\rule{0ex}{0ex}}orMgh={v}^{2}\left(\frac{1}{2}M+\frac{1}{2}\frac{I}{{r}^{2}}\right)\phantom{\rule{0ex}{0ex}}or{v}^{2}=\frac{Mgh}{\left(\frac{1}{2}M+\frac{1}{2}\frac{I}{{r}^{2}}\right)}................\left(1\right)$Step5: Finding the accelerationWe know from the formulae of kinematics that, ${v}^{2}={u}^{2}-2as$, where s is the displacement of the body, a is acceleration, and u and v are the initial and final velocity.In this case, the body starts to move from rest.So, ${v}^{2}=2as..............\left(2\right)$From 1 and 2$2as=\frac{Mgh}{\left(\frac{1}{2}M+\frac{1}{2}\frac{I}{{r}^{2}}\right)}=\frac{2gh}{\left(1+\frac{1}{M{r}^{2}}\right)}\phantom{\rule{0ex}{0ex}}ora=\frac{\frac{2gh}{\left(1+\frac{1}{M{r}^{2}}\right)}}{2s}=\frac{\frac{2gh}{\left(1+\frac{1}{M{r}^{2}}\right)}}{\frac{2h}{\mathrm{sin}\theta }}=\frac{g\mathrm{sin}\theta }{\left(1+\frac{1}{M{r}^{2}}\right)}\left(fromfigure,\mathrm{sin}\theta =\frac{h}{s}\right)\phantom{\rule{0ex}{0ex}}ora=\frac{g\mathrm{sin}\theta }{\left(1+\frac{1}{M{r}^{2}}\right)}$Step6: Diagram Therefore, the acceleration of the body is $a=\frac{g\mathrm{sin}\theta }{\left(1+\frac{1}{M{r}^{2}}\right)}$.  Suggest Corrections  0      Similar questions  Explore more