91Ó°ÊÓ

It is given that the mass of each object as $M=0.840\mathrm{Kg}$, ramp inclination as $\mathrm{θ}=35.0°$and initial height as h.

Given that all the objects have same mass and height that implies the gravitational potential energy is also same but the moment of inertia is not the same.

The moment of inertia of the given shapes are:

Uniform solid cylinder: $l_{usc}=\frac{1}{2}M{R}^2$, Hollow cylinder, $I_{hc}=M{R}^2$, uniform solid sphere $l_{uss}=\frac{2}{5}M{R}^2$, hollow sphere $l_{hs}=\frac{2}{3}M{R}^2$.

Here, the initial energy is $Mgh$and the final energy is the sum of the kinetic energy which can be written as $Mgh=K_{cm}+K_{rot}$…… (1)

Substitute $K_{cm}=\frac{1}{2}M{v}_{cm}^2$and $K_{rot}=\frac{1}{2}l{\mathrm{Ó\neg }}^2$ in (1) then, $Mgh=\frac{1}{2}M{v}_{cm}^2+\frac{1}{2}l{\mathrm{Ó\neg }}^2$…… (2)

The speed of center of mass is $\mathrm{Ó\neg }=\frac{v_{cm}}{R}$. Since, all the moment of inertia is proportional to $M{R}^2$, $l$can be written as $l=\mathrm{η}M{R}^2$ where $\mathrm{η}$ is a constant.

Plug all the known values in (2) and simplify:

$$\begin{gathered} Mgh=\frac{1}{2}M{v}_{cm}^2+\frac{1}{2}\left(\mathrm{η}M{R}^2\right){\left(\frac{v_{cm}}{R}\right)}^2 \\ =\frac{1}{2}M{v}_{cm}^2+\frac{1}{2}\mathrm{η}MR{v}_{cm}^2 \\ ⇒gh=\frac{1}{2}{v}_{cm}^2+\frac{1}{2}\mathrm{η}R{v}_{cm}^2 \\ ⇒gh=\frac{v_{cm}^2}{2}(1+\mathrm{η}R) \end{gathered}$$

From the above equation,$v_{cm}=\sqrt{\frac{2gh}{1+\mathrm{η}}}$ . The constant in the moment of inertia ranges as ${\mathrm{η}}_{uss}<{\mathrm{η}}_{usc}<{\mathrm{η}}_{hs}<{\mathrm{η}}_{hc}$. Thus, the uniform solid sphere arrives first, uniform solid arrives second, hollows sphere third, hollow cylinder arrives fourth.

Here, all the objects have same potential energy at the start. Then depends on the conservation of energy all objects have same kinetic energy at the bottom.

The rotational kinetic energy at the bottom is $K_{rot}=\frac{1}{2}l{\mathrm{Ó\neg }}^2$…… (3)

Substitute $l=\mathrm{η}M{R}^2,\mathrm{Ó\neg }=\frac{v_{cm}}{R},v_{cm}=\sqrt{\frac{2gh}{1+\mathrm{η}}}$ in (3) and simplify.

$$\begin{gathered} K_{rot}=\frac{1}{2}\left(\mathrm{η}M{R}^2\right){\left(\frac{v_{cm}}{R}\right)}^2 \\ =\frac{\mathrm{η}MR{v}_{cm}^2}{2} \\ =\frac{\mathrm{η}MR}{2}\left(\frac{2gh}{1+\mathrm{η}}\right) \\ =\frac{\mathrm{η}MRgh}{1+\mathrm{η}} \end{gathered}$$

Therefore, the object with largest constant $\mathrm{η}$has the greatest kinetic energy at the bottom. Here, hollow cylinder has the greatest kinetic energy.

If $v_{cm}=\mathrm{Ó\neg }R$ and $a_{cm}=\mathrm{Î\pm }R$ then the object of R does not slip.

Here, the frictional force is constant and is equal to localid="1667996845926" $F_{fr}=\mathrm{μ}Mg\cos \mathrm{θ}$and the torque produced by the force is $\mathrm{τ}=R\mathrm{μ}Mg\cos \mathrm{θ}$.

Since, $\mathrm{Ï„}=l\mathrm{Î\pm }$ then, $l\mathrm{Î\pm }=R\mathrm{μ}Mg\cos \mathrm{θ}$and $\mathrm{μ}=\frac{l\mathrm{Î\pm }}{RMg\cos \mathrm{θ}}$.

Substitute $\mathrm{Ï„}=\mathrm{η}M{R}^2$ and $a_{cm}=\mathrm{Î\pm }R$ in $\mathrm{μ}$.

$$\begin{gathered} \mathrm{μ}=\frac{\mathrm{η}M{R}^2\left(\frac{a_{om}}{R}\right)}{RMg\cos \mathrm{θ}} \\ =\frac{\mathrm{η}{a}_{cm}}{g\cos \mathrm{θ}} \end{gathered}$$

By Newton’s second law, the acceleration $a_{cm}$ can be written as follows:

$$\begin{gathered} M{a}_{an}=Mg\sin \mathrm{θ}−\mathrm{μ}Mg\cos \mathrm{θ} \\ ⇒a_{cm}=g\sin \mathrm{θ}−\mathrm{μ}g\cos \mathrm{θ} \\ ⇒\mathrm{μ}g\cos \mathrm{θ}=g\sin \mathrm{θ}−a_{cm} \\ ⇒\mathrm{μ}=\frac{g\sin \mathrm{θ}−a_{cm}}{g\cos \mathrm{θ}} \end{gathered}$$

Further, the above equation can be written as follows:

$$\begin{gathered} \mathrm{μ}=\frac{\mathrm{η}g\sin \mathrm{θ}−\mathrm{η}g\mathrm{μ}\cos \mathrm{θ}}{g\cos \mathrm{θ}} \\ \mathrm{μ}=\frac{\mathrm{η}g(\sin \mathrm{θ}−\mathrm{μ}\cos \mathrm{θ})}{g\cos \mathrm{θ}} \\ \mathrm{μ}\cos \mathrm{θ}=\mathrm{η}\sin \mathrm{θ}−\mathrm{μ}\mathrm{κ}\cos \mathrm{θ} \\ \mathrm{μ}\cos \mathrm{θ}(1+\mathrm{κ})=\mathrm{η}\sin \mathrm{θ} \end{gathered}$$

From the above equation, $\mathrm{μ}=\frac{\tan \mathrm{θ}}{\left(1+\mathrm{κ}\right)}$. Substitute $\mathrm{κ}=1$and $\mathrm{θ}=35.0°$ then,

$$\begin{gathered} \mathrm{μ}=\frac{\tan \mathrm{θ}}{(1+\mathrm{κ})} \\ =\frac{\tan \left(35.0\right)}{(1+1)} \\ =0.35 \end{gathered}$$

Therefore, if the coefficient of friction is greater than or equal to 0.35 the object will roll without slipping.