Question 406: Centre of Mass and Rotational Motion

Question

A solid sphere and a hollow cylinder roll up without slipping on same inclined plane with same initial speed $$v$$. The sphere and the cylinder reaches upto maximum heights $$h_1$$ and $$h_2$$ respectively, above the initial level. The ratio $$h_1: h_2$$ is $$\frac{n}{10}$$. The value of $$n$$ is __________.

Accepted Answer

To solve this problem, we first note that for both the solid sphere and the hollow cylinder, the total mechanical energy is conserved as they roll up the inclined plane without slipping. The initial kinetic energy (comprised of both translational and rotational kinetic energy) is converted into potential energy at the maximum height.
Kinetic Energy for Each Body at the Start:
For the solid sphere, the moment of inertia $$I$$ is given by $$I = \frac{2}{5}mr^2$$, where $$m$$ is mass and $$r$$ is the radius of the sphere. The kinetic energy is the sum of translational kinetic energy $$\left(\frac{1}{2}mv^2ight)$$ and rotational kinetic energy $$\left(\frac{1}{2}I\omega^2ight)$$, where $$\omega$$ is the angular velocity. Since the sphere rolls without slipping, $$v = r\omega$$.
The total initial kinetic energy for the solid sphere is:
$$KE_{	ext{sphere}} = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2ight)\left(\frac{v}{r}ight)^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$$
For the hollow cylinder, the moment of inertia $$I$$ is $$mr^2$$. Thus, its total kinetic energy is:
$$KE_{	ext{cylinder}} = \frac{1}{2}mv^2 + \frac{1}{2}mr^2\left(\frac{v}{r}ight)^2 = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$$
Potential Energy at Maximum Height:
For both bodies, the potential energy at the maximum height is given by $$PE = mgh$$, where $$h$$ is the height reached.
Applying Conservation of Energy:
For the solid sphere, the energy conservation equation is:
$$\frac{7}{10}mv^2 = mgh_1$$
Solving for $$h_1$$ gives:
$$h_1 = \frac{7v^2}{10g}$$
For the hollow cylinder, the conservation of energy gives:
$$mv^2 = mgh_2$$
Thus, $$h_2 = \frac{v^2}{g}$$.
Finding the Ratio $$h_1:h_2$$:
The ratio of $$h_1$$ to $$h_2$$ is:
$$\frac{h_1}{h_2} = \frac{\frac{7v^2}{10g}}{\frac{v^2}{g}} = \frac{7}{10}$$
Therefore, the value of $$n$$, which represents the numerator in the ratio $$\frac{n}{10}$$, is $$7$$. Thus, the ratio of maximum heights $$h_1:h_2$$ reached by the solid sphere and the hollow cylinder, respectively, is $$\frac{7}{10}$$.

Centre of Mass and Rotational Motion

A solid sphere and a hollow cylinder roll up without slipping on same inclined plane with same initial speed $v. The sphere and the cylinder reaches upto maximum heights h_1 and h_2 respectively, above the initial level. The ratio h_1: h_2 is \frac{n}{10}. The value of n$ is __________.

Integer / Numeric Answer Type