A thin light string is wrapped around the outer rim of a uniform hollow cylinder of mass $\mathbf{4}\mathbf{.}\mathbf{75}\mathbf{}\mathbf{kg}$having inner and outer radii as shown in Fig. $\mathbf{E}\mathbf{10}\mathbf{.}\mathbf{25}$. the cylinder is then released from rest. (a) How far must the cylinder fall before its center is moving at $\mathbf{6}\mathbf{.}\mathbf{66}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$? (b) If you just dropped this cylinder without any string, how fast would its center be moving when it had fallen the distance in part (a)? (c) Why do you get two different answers when the cylinder falls the same distance in both cases?

Mass of hollow cylinder $\left(m\right)$=4.75 kg.

The inner radius of the cylinder $\left(r_1\right)$localid="1667971746016" $=20.00\mathrm{cm}=0.200\mathrm{m}$.

The outer radius of the cylinder localid="1667971737711" $\left({\mathrm{r}}_2\right)=35.0\mathrm{cm}=0.350\mathrm{cm}$.

The cylinder is released from the rest.

The sum of the initial energies of the system plus work done on the system by external forces is equal to the sum of the final energies of the system.

$E_i+W=E_f \cdots \cdots \left(1\right)$

The gravitational potential energy of the system is given by,

$U=mgh \cdots \cdots \left(2\right)$

The total kinetic energy of the rigid object rolling on a rough surface without slipping is the sum of the rotational kinetic energy about its center of mass and the translational kinetic energy of the center of mass.

$$\begin{gathered} K_{tot}=K_{tr}+K_{rot} \\ K_{tot}=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2 \cdots \cdots \left(3\right) \end{gathered}$$

The velocity of center of mass of of the rigid object rolling on a rough surface without slipping can be expressed as:

$v=R\omega \cdots \cdots \left(4\right)$

where, $\omega$is the angular speed of the object about its center of mass and$R$is radius of the rigid object.

Consider the system as the cylinder and the earth.

Let the initial state of the system be at the point from where it is released and the final state be the point at which the center of the cylinder is moving at $6.66\mathrm{m}/\mathrm{s}.$

Let the zero-gravitational potential energy of the system be at the final state.

Now, from equation$\left(1\right)$, applying conservation of energy principle to the system of cylinder and the Earth between the initial and final state as:

$E_i+W=E_f$, where,

$E_i$is the initial energy of the system which is the sum of the initial kinetic energy and gravitational potential energy, $W=0$, since there are no external forces acting on the system, and $E_f$is the final energy of the system which is the sum of the final kinetic energy and gravitational potential energy.

Therefore,

$K_i+U_{gi}+0=K_f+U_{gf} \cdots \cdots \left(5\right)$.

We have

$K_i=0$, since the system starts at the rest in the initial state and $U_{gf}=0$.

Thus, using $\left(3\right)$in above equation we get,

localid="1664282266218" $$\begin{gathered} 0+mgh+0=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2+0 \\ \Rightarrow mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2 \end{gathered}$$

Since we have moment if inertia $I=\frac{1}{2}m\left(r_1^2+r_2^2\right)$, substituting in above equation, we get,

$$\begin{gathered} mgh=\frac{1}{2}m{v}^2+\frac{1}{2}·\frac{1}{2}m\left(r_1^2+r_2^2\right){\omega }^2 \\ \Rightarrow gh=\frac{1}{2}{v}^2+\frac{1}{4}\left(r_1^2+r_2^2\right){\omega }^2 \end{gathered}$$

Substituting the value from $\left(4\right)$in above equation, we get,

width="232">gh=12v2+14r12+r22vr22⇒h=12gv2+14gr12+r22vr22⇒h=12gv2+12r12+r22vr22

$$\begin{gathered} \Rightarrow h=\frac{1}{2\left(9.80\right)}\left({\left(6.66\right)}^2+\frac{1}{2}\left(\left(0.200\right)+{\left(0.350\right)}^2\right){\left(\frac{6.66}{0.350}\right)}^2\right) \\ \Rightarrow h=3.76 m \end{gathered}$$

Hence, the height at which the cylinder fall is, $h=3.76 m$.

Here, the final kinetic energy would only be the translational kinetic energy of the cylinder’s center of mass.

From equation $\left(5\right)$becomes,

$$\begin{gathered} mgh=\frac{{\displaystyle 1}}{{\displaystyle 2}}m{v}^2 \\ \Rightarrow gh=\frac{{\displaystyle 1}}{{\displaystyle 2}}{v}^2 \\ \Rightarrow v=\sqrt{2gh} \\ \Rightarrow v=\sqrt{2\left(9.80\right)·\left(3.76\right)} \\ \Rightarrow v=8.58 m/s. \end{gathered}$$

Hence, the velocity of the is $v=8.58 m/s.$

The answers are different because in part (a), the gravitational potential energy is converted into two types of kinetic energies, rotational and translational kinetic energy.

In part (b), the gravitational potential energy is converted to only translational kinetic energy.