# A solid cylinder rolls down an incline plane without slipping. If the center of mass of the...

## Question:

A solid cylinder rolls down an incline plane without slipping. If the center of mass of the cylinder has a linear acceleration of 2.67 {eq}m/s^2 {/eq}, what is the angle of the incline to the horizontal?

## Conservation of energy

Whenever the work is done on a body, it will gain some energy, and will remain the same but changes to another forms of energy like the change of potential energy to rotational energy

Given data

• Linear acceleration of the cylinder {eq}\left( a \right) = 2.67\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} {/eq}
• Initial velocity {eq}\left( u \right) = 0 {/eq}
• final velocity {eq}\left( v \right) {/eq}
• Height from which the cylinder is rolling {eq}\left( h \right) {/eq}
• Distance travelled by the cylinder {eq}\left( S \right) {/eq}

Since the cylinder rolls down in an inclined plane, the potential energy is converted into kinetic energy as well as rotational energy.

The expression for the transitional energy is

{eq}{E_t} = K{E_t} = \dfrac{1}{2}m{v^2}.........\left( 1 \right) {/eq}

The expression for the rotational kinetic energy is

{eq}{E_r} = K{E_r} = \dfrac{1}{2}I{\omega ^2}...........\left( 2 \right) {/eq}

The expression for the moment of inertia is

{eq}I = \dfrac{1}{2}m{r^2}.........\left( 3 \right) {/eq}

Now the expression for the conservation of energy is

{eq}PE = {E_r} + {E_t}...........\left( 4 \right) {/eq}

Substituting the values in above expression we get

{eq}\begin{align*} mgh &= \left[ {\dfrac{1}{2}\left( {\dfrac{1}{2}m{r^2}} \right){\omega ^2}} \right] + \left( {\dfrac{1}{2}m{v^2}} \right)\\ mgh &= \left[ {\dfrac{1}{2}\left( {\dfrac{1}{2}m{r^2}} \right){{\left( {\dfrac{v}{r}} \right)}^2}} \right] + \left( {\dfrac{1}{2}m{v^2}} \right)\\ mgh &= \dfrac{1}{4}m{v^2} + \dfrac{1}{2}m{v^2}\\ \sqrt {\dfrac{4}{3}gh} &= v.........\left( 5 \right) \end{align*} {/eq}

The expression for the third equation of motion is

{eq}{v^2} = {u^2} + 2aS {/eq}

Substituting the values in above expression we get

{eq}\begin{align*} {\left( {\sqrt {\dfrac{4}{3}gh} } \right)^2} &= 0 + 2aS\\ \dfrac{4}{3}gh &= 2aS\\ \dfrac{h}{S} &= \dfrac{{3a}}{{2g}}...........\left( 6 \right) \end{align*} {/eq}

Substituting the values in above expression 6 we get

{eq}\begin{align*} \dfrac{h}{S} &= \dfrac{{3 \times 2.67}}{{2 \times 9.81}}\\ \dfrac{h}{S} &= 0.408 \end{align*} {/eq}

Since for the inclined plane {eq}\sin \theta = \dfrac{h}{S} {/eq}

Therefore

{eq}\begin{align*} \sin \theta &= 0.408\\ \theta &= 24.07^\circ \end{align*} {/eq}

therefore the angle of the plane is {eq}\theta = 24.07^\circ {/eq}