You have one object of each of these shapes, all with mass 0.840 kg: a uniform solid cylinder, a thin-walled hollow cylinder, a uniform solid sphere, and a thin-walled hollow sphere. You release each object from rest at the same vertical height h above the bottom of a long wooden ramp that is inclined at $35.0°$ from the horizontal. Each object rolls without slipping down the ramp. You measure the time t that it takes each one to reach the bottom of the ramp; Fig. P10.89 shows the results. (a) From the bar graphs, identify objects A through D by shape. (b) Which of objects A through D has the greatest total kinetic energy at the bottom of the ramp, or do all have the same kinetic energy? (c) Which of objects A through D has the greatest rotational kinetic energy$\frac{1}{2}l{\omega }^2$ at the bottom of the ramp, or do all have the same rotational kinetic energy? (d) What minimum coefficient of static friction is required for all four objects to roll without slipping?

It is given that the mass of each object as $M=0.840\mathrm{Kg}$, ramp inclination as $\theta =35.0°$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 $K_{rot}=\frac{1}{2}l{\omega }^2$ in (1) then, $Mgh=\frac{1}{2}M{v}_{cm}^2+\frac{1}{2}l{\omega }^2$…… (2)

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

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

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

From the above equation,$v_{cm}=\sqrt{\frac{2gh}{1+\eta }}$ . The constant in the moment of inertia ranges as ${\eta }_{uss}<{\eta }_{usc}<{\eta }_{hs}<{\eta }_{hc}$. 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 $K_{rot}=\frac{1}{2}l{\omega }^2$…… (3)

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

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

Therefore, the object with largest constant $\eta$has the greatest kinetic energy at the bottom. Here, hollow cylinder has the greatest kinetic energy.

If $v_{cm}=\omega R$ and $a_{cm}=\alpha R$ then the object of R does not slip.

Here, the frictional force is constant and is equal to localid="1667996845926" $F_{fr}=\mu Mg\cos \theta$and the torque produced by the force is $\tau =R\mu Mg\cos \theta$.

Since, $\tau =l\alpha$ then, $l\alpha =R\mu Mg\cos \theta$and $\mu =\frac{l\alpha }{RMg\cos \theta }$.

Substitute $\tau =\eta M{R}^2$ and $a_{cm}=\alpha R$ in $\mu$.

$$\begin{gathered} \mu =\frac{\eta M{R}^2\left(\frac{a_{om}}{R}\right)}{RMg\cos \theta } \\ =\frac{\eta a_{cm}}{g\cos \theta } \end{gathered}$$

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

$$\begin{gathered} M{a}_{an}=Mg\sin \theta -\mu Mg\cos \theta \\ \Rightarrow a_{cm}=g\sin \theta -\mu g\cos \theta \\ \Rightarrow \mu g\cos \theta =g\sin \theta -a_{cm} \\ \Rightarrow \mu =\frac{g\sin \theta -a_{cm}}{g\cos \theta } \end{gathered}$$

Further, the above equation can be written as follows:

$$\begin{gathered} \mu =\frac{\eta g\sin \theta -\eta g\mu \cos \theta }{g\cos \theta } \\ \mu =\frac{\eta g(\sin \theta -\mu \cos \theta )}{g\cos \theta } \\ \mu \cos \theta =\eta \sin \theta -\mu \kappa \cos \theta \\ \mu \cos \theta (1+\kappa )=\eta \sin \theta \end{gathered}$$

From the above equation, $\mu =\frac{\tan \theta }{\left(1+\kappa \right)}$. Substitute $\kappa =1$and $\theta =35.0°$ then,

$$\begin{gathered} \mu =\frac{\tan \theta }{(1+\kappa )} \\ =\frac{\tan \left(35.0\right)}{(1+1)} \\ =0.35 \end{gathered}$$

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