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Step-by-Step Solution
Step 1: Write the Total Kinetic Energy of a Rolling Body
The total kinetic energy of a rolling body (rolling without slipping) is given by:
$$
E_\text{total} = \frac{1}{2} m v^2 \Bigl[1 + \frac{K^2}{R^2}\Bigr]
$$
Here,
- $m$ is the mass of the body,
- $v$ is the speed of the center of mass,
- $R$ is the radius of the rolling object, and
- $K$ is the radius of gyration for that rolling object.
Step 2: Apply Conservation of Energy
As the body rolls up the incline, its kinetic energy is converted into potential energy at some maximum height $h$. Hence:
$$
\frac{1}{2} m v^2 \Bigl[1 + \frac{K^2}{R^2}\Bigr] = m g h
$$
Solving for $h$ gives:
$$
h = \frac{v^2}{2g} \Bigl[1 + \frac{K^2}{R^2}\Bigr]
$$
Step 3: Calculate the Height for Each Object
For the ring:
The radius of gyration for a ring of radius $R$ is $K = R$, so
$$
\frac{K^2}{R^2} = 1.
$$
Hence,
$$
h_1 = \frac{v^2}{2g} \Bigl[1 + 1\Bigr] = \frac{v^2}{g}.
$$
For the solid cylinder:
A solid cylinder of radius $R/2$ has radius of gyration $K = \frac{R}{2\sqrt{2}}$. Thus,
$$
\frac{K^2}{(R/2)^2}
= \frac{\left(\frac{R}{2\sqrt{2}}\right)^2}{\left(\frac{R}{2}\right)^2}
= \frac{\frac{R^2}{8}}{\frac{R^2}{4}}
= \frac{1}{2}.
$$
So,
$$
h_2 = \frac{v^2}{2g} \Bigl[1 + \frac{1}{2}\Bigr]
= \frac{3v^2}{4g}.
$$
For the solid sphere:
A solid sphere of radius $R/4$ has a moment of inertia given by
$I = \frac{2}{5} m \bigl(\frac{R}{4}\bigr)^2$; effectively, the ratio
$$
\frac{K^2}{R^2} = \frac{2}{5}.
$$
Therefore,
$$
h_3 = \frac{v^2}{2g} \Bigl[1 + \frac{2}{5}\Bigr]
= \frac{7v^2}{10g}.
$$
Step 4: Determine the Ratio of Heights Climbed
The ratio of the heights for the ring ($h_1$), solid cylinder ($h_2$), and solid sphere ($h_3$) is:
$$
h_1 : h_2 : h_3
= \frac{v^2}{g} : \frac{3v^2}{4g} : \frac{7v^2}{10g}.
$$
Simplifying each term to a common base, we get:
$$
1 : \frac{3}{4} : \frac{7}{10} = 20 : 15 : 14.
$$
Hence, the required ratio of maximum heights is 20 : 15 : 14.
