Chapter 8, Problem 50P

Four objects—a hoop, a solid cylinder, a solid sphere, and a thin, spherical shell—each have a mass of 4.80 kg and a radius of 0.230 m. (a) Find the moment of inertia for each object as it rotates about the axes shown in Table 8.1. (b) Suppose each object is rolled down a ramp. Rank the translational speed of each object from highest to lowest, (c) Rank the objects’ rotational kinetic energies from highest to lowest as the objects roll down the ramp.

To determine

The moment of inertia of the each of the object it rotates.

Answer to Problem 50P

The moment of inertia of the each of the object it rotates is, hoop is $0.254{\text{kgm}}^{\text{2}}$, solid cylinder is $0.127{\text{kgm}}^{\text{2}}$, solid sphere is $0.102{\text{kgm}}^{\text{2}}$, thin spherical shell is $0.169{\text{kgm}}^{\text{2}}$.

Explanation of Solution

Given Info: mass of the hoop $m_{\text{h}}$ is $4.80\text{kg}$  and radius of the hoop $r_{\text{h}}$ is $0.230{\text{m}}^{\text{2}}$

Formula to calculate the moment of inertia of the hoop,

$I_{\text{h}}=m_{\text{h}}{r}_{\text{h}}{}^2$

• $I_{\text{h}}$ is the moment of inertia of the hoop,
• $m_{\text{h}}$ is the mass of the hoop,
• $r_{\text{h}}$ is the radius of the hoop,

Substitute $4.80\text{kg}$ for $m_{\text{h}}$ and $0.230{\text{m}}^{\text{2}}$ for $r_{\text{h}}$ to find $I_{\text{h}}$,

$\begin{array}{c}{I}_{\text{h}}=\left(4.80\text{kg}\right){\left(0.230{\text{m}}^{\text{2}}\right)}^2\\ =\left(4.80\text{kg}\right)\left(0.0529{\text{m}}^{\text{2}}\right)\\ =0.2539{\text{kgm}}^{\text{2}}\\ \approx 0.254{\text{kgm}}^{\text{2}}\end{array}$

The moment of inertia of the hoop is $0.254{\text{kgm}}^{\text{2}}$

Formula to calculate the moment of inertia of the solid cylinder,

$I_{\text{sc}}=\frac{1}{2}{m}_{\text{sc}}{r}_{\text{sc}}{}^2$

• $I_{\text{sc}}$ is the moment of inertia of the solid cylinder,
• $m_{\text{sc}}$ is the mass of the solid cylinder,
• $r_{\text{sc}}$ is the radius of the solid cylinder,

Substitute $4.80\text{kg}$ for $m_{\text{sc}}$ and $0.230{\text{m}}^{\text{2}}$ for $r_{\text{sc}}$ to find $I_{\text{sc}}$,

$\begin{array}{c}{I}_{\text{sc}}=\frac{1}{2}\left[\left(4.80\text{kg}\right){\left(0.230{\text{m}}^{\text{2}}\right)}^2\right]\\ =\frac{1}{2}\left[\left(4.80\text{kg}\right)\left(0.0529{\text{m}}^{\text{2}}\right)\right]\\ =0.1179{\text{kgm}}^{\text{2}}\\ \approx 0.127{\text{kgm}}^{\text{2}}\end{array}$

The moment of inertia of the solid cylinder is $0.127{\text{kgm}}^{\text{2}}$.

Formula to calculate the moment of inertia of the solid sphere,

$I_{\text{ss}}=\frac{1}{2}{m}_{\text{ss}}{r}_{\text{ss}}{}^2$

• $I_{\text{ss}}$ is the moment of inertia of the solid sphere,
• $m_{\text{ss}}$ is the mass of the solid sphere,
• $r_{\text{ss}}$ is the radius of the solid sphere,

Substitute $4.80\text{kg}$ for $m_{\text{ss}}$ and $0.230{\text{m}}^{\text{2}}$ for $r_{\text{ss}}$ to find $I_{\text{ss}}$,

$\begin{array}{c}{I}_{\text{ss}}=\frac{2}{5}\left[\left(4.80\text{kg}\right){\left(0.230{\text{m}}^{\text{2}}\right)}^2\right]\\ =\frac{2}{5}\left[\left(4.80\text{kg}\right)\left(0.0529{\text{m}}^{\text{2}}\right)\right]\\ =0.1015{\text{kgm}}^{\text{2}}\\ \approx 0.102{\text{kgm}}^{\text{2}}\end{array}$

Thus, the moment of inertia of the solid sphere is $0.102{\text{kgm}}^{\text{2}}$

Formula to calculate the moment of inertia of the thin spherical shell,

$I_{\text{tss}}=\frac{1}{2}{m}_{\text{tss}}{r}_{\text{tss}}{}^2$

• $I_{\text{tss}}$ is the moment of inertia of the thin spherical shell,
• $m_{\text{tss}}$ is the mass of the thin spherical shell,
• $r_{\text{tss}}$ is the radius of the thin spherical shell,

Substitute $4.80\text{kg}$ for $m_{\text{tss}}$ and $0.230\text{m}$ for $r_{\text{tss}}$ to find $I_{\text{tss}}$,

$\begin{array}{c}{I}_{\text{ss}}=\frac{2}{3}\left[\left(4.80\text{kg}\right){\left(0.230{\text{m}}^{\text{2}}\right)}^2\right]\\ =\frac{2}{3}\left[\left(4.80\text{kg}\right)\left(0.0529{\text{m}}^{\text{2}}\right)\right]\\ =0.169{\text{kgm}}^{\text{2}}\end{array}$

The moment of inertia of the thin spherical shell is $0.169{\text{kgm}}^{\text{2}}$

Conclusion:

Therefore, the moment of inertia of the each of the object it rotates is, hoop is $0.254{\text{kgm}}^{\text{2}}$, solid cylinder is $0.127{\text{kgm}}^{\text{2}}$, solid sphere is $0.102{\text{kgm}}^{\text{2}}$, thin spherical shell is $0.169{\text{kgm}}^{\text{2}}$.

To determine

The translational speed of each of the object from highest to lowest when it rolled down the ramp.

Answer to Problem 50P

The translational speed of each of the object from highest to lowest when it rolled down the ramp is Solid sphere, solid cylinder, thin spherical shell and hoop.

Explanation of Solution

Given Info: M, R, g and $\theta$ all have same values for all the objects.

Explanation:

Formula to calculate translational acceleration is,

$a=R\alpha$

• a is the translational acceleration,
• R is the radius,
• $\alpha$ is the angular acceleration.

Rearrange in terms of $\alpha$.

$\alpha =\frac{a}{R}$

The below figure shows the forces acting on the object.

From Newton’s second law,

$\Sigma F_X=Ma$

Consider the force diagram,

$\Sigma F_X=Mg\sin \theta -f$

Equate both above force equations,

$Ma=Mg\sin \theta -f$

Formula to calculate the torque is,

$\tau =I\alpha$

Formula to calculate the torque in terms of f is,

$\tau =fR$

Equate (I) and (II)

$\begin{array}{l}fR=I\alpha \\ f=\frac{I\alpha }{R}\end{array}$

Use $a/R$ for $\alpha$ in the above equation to rewrite f.

$\begin{array}{c}f=\frac{I\left(a/R\right)}{R}\\ =\frac{Ia}{R^2}\end{array}$

Substitute $Ia/R^2$ for f in the equation $Ma=Mg\sin \theta -f$ to find a,

$\begin{array}{c}Ma=Mg\sin \theta -\frac{Ia}{R^2}\\ a=\frac{Mg\sin \theta }{M+I/R^2}\end{array}$

Since M, R, g and $\theta$ all have same values for all the objects.

From the above equation, the translational acceleration increases as the value of the moment of inertia decreases.

Conclusion:

The translational speed of each of the object from highest to lowest when it rolled down the ramp is Solid sphere, solid cylinder, thin spherical shell and hoop.

To determine

The rotational kinetic energy of each of the object from highest to lowest when it rolled down the ramp.

Answer to Problem 50P

The rotational kinetic energy of each of the object from highest to lowest when it rolled down the ramp is Hoop, thin spherical shell, solid cylinder , and Solid sphere,

Explanation of Solution

Given Info: M, g and h all have same values for all the objects.

Explanation:

Apply principle of conservation of energy. Potential energy of the object at the top of the incline is equal to the rotational and translational kinetic energy of the object at the bottom of the incline.

$K{E}_r+K{E}_t=\Delta P{E}_g$

• $K{E}_r$ is the rotational kinetic energy,
• $K{E}_t$ is the transitional kinetic energy,
• $\Delta P{E}_g$ is the gravitational potential energy.

When the objects roll down frictional force is zero in it and the Transitional kinetic energy is conserved

Use $\frac{1}{2}M{v}^2$ for $K{E}_t$ and $Mgh$ for $\Delta P{E}_g$ in the above equation,

$K{E}_r=Mgh-\frac{1}{2}M{v}^2$

• h is the vertical height of the object from the ground at the top of the ramp
• v is the translational speed.

M, g and h all have same values for all the objects

From the above equation the rotational kinetic energy decreases as the value of the translational speed increase.

Conclusion:

The rotational kinetic energy of each of the object from highest to lowest when it rolled down the ramp is Hoop, thin spherical shell, solid cylinder , and Solid sphere,