Question

In Fig. 15-59, a solid cylinder attached to a horizontal spring (k=3.00 N/m) rolls without slipping along a horizontal surface. If the system is released from rest when the spring is stretched by 0.250 m , find (a) the translational kinetic energy and (b) the rotational kinetic energy of the cylinder as it passes through the equilibrium position. (c) Show that under these conditions the cylinder’s center of mass executes simple harmonic motion with period $\mathit{T}\mathbf{=}\mathbf{2}\mathbf{\pi }\sqrt{\frac{\mathbf{3}\mathbf{M}}{\mathbf{2}\mathbf{k}}}$ where M is the cylinder mass. (Hint: Find the time derivative of the total mechanical energy.)

Short answer

1. The translational K.E of the cylinder as it passes through the equilibrium position is 0.0625 J.
2. The rotational K.E of the cylinder as it passes through the equilibrium position is $3.13\times {10}^{-2}\mathrm{J}$.
3. The period of the SHM executed by the cylinder’s center of mass is $T=2\tau \tau \sqrt{\frac{3M}{2k}}$.

Step-by-step solution

The given data

• The spring constant is, k=3.0 N/m
• The displacement of the spring is, $x_m=0.250m$
• The solid cylinder rolls on the horizontal surface without slipping.

Understanding the concept of SHM

The energy of a body by virtue of its translational motion, is known as Translational kinetic energy and the energy of the body due to its rotational motion is called rotational kinetic energy. The law of conservation of energy states that the maximum potential energy of the body should be equal to the maximum value of the sum of translational and rotational kinetic energy. Then using the condition for total energy to be constant, we can find the given expression for the period of SHM executed by the cylinder’s center of mass.

Formulae:

The translational kinetic energy of the system, $K_T=\frac{1}{2}M{v}^2$ (i)

Here, M is mass, v is velocity of the object.

Rotational kinetic energy of the system, $K_R=\frac{1}{2}l{w}^2$ (ii)

Here, l is rotational inertia, $\omega$is angular velocity.

Moment of inertia of cylinder through its center and perpendicular to its plane,

$l=\frac{1}{2}M{R}^2$ (iii)

Here, l is the rotational inertia, M is mass, R is the radius of the cylinder.

The period of oscillation for SHM,

$T=2\tau \tau \sqrt{\frac{M}{k}}$ (iv)

Here, T is the time period, M is mass, k is the force constant.

The angular speed of the system, $\omega =\frac{v}{R}$ (v)

Here, v is the velocity of the object, $\omega$is angular velocity and R is the radius of the cylinder.

a) Calculation of translational kinetic energy

According to the law of conservation of energy, total energy content of the cylinder (E) is constant . So, for the center of mass(cm) , we get the equation-

$$\begin{gathered} \frac{1}{2}k{x}_m^2=\frac{1}{2}M{v}_{cm}^2+\frac{1}{2}{l}_{cm}{\omega }^2 \\ \frac{1}{2}k{x}_m^2=\frac{1}{2}M{v}_{cm}^2+\frac{1}{2}\left(\frac{1}{2}M{R}^2\right){\omega }^2 \\ \frac{1}{2}k{x}_m^2=\frac{1}{2}M{v}_{cm}^2+\frac{1}{4}M{v}_{cm}^2 \\ \frac{1}{2}k{x}_m^2=\frac{3}{4}M{v}_{cm}^2 \end{gathered}$$

On further solving

$$\begin{gathered} {\mathrm{Mv}}^2=\frac{2}{3}{\mathrm{kx}}_{\mathrm{m}}^2 \\ =\frac{2}{3}\left(3\mathrm{N}/\mathrm{m}\right){\left(0.25\mathrm{m}\right)}^2 \\ =0.125\mathrm{J} \end{gathered}$$

Therefore, the translational K.E of the cylinder as it passes through the equilibrium position from equation (i) is given as:

$$\begin{gathered} \mathrm{K}.{\mathrm{E}}_{.\mathrm{trans}}=\frac{1}{2}{\mathrm{Mv}}^2 \\ =\frac{1}{2}\times 0.125\mathrm{J} \\ =0.0625\mathrm{J} \end{gathered}$$

Hence, the value of translational K.E. is 0.0625 J.

b) Calculation of rotational kinetic energy

From part (a), in the derivational it can be seen that the value of rotational energy is given as:

$$\begin{gathered} \frac{1}{2}{\mathrm{l}}_{\mathrm{cm}}{\mathrm{\omega }}^2=\frac{1}{4}{\mathrm{Mv}}_{\mathrm{cm}}^2 \\ =\frac{1}{4}\left(0.125\right) \\ =3.13\times {10}^{-2}\mathrm{J} \end{gathered}$$

Therefore, the rotational K.E of the cylinder as it passes through the equilibrium position is $3.13\times {10}^{-2}\mathrm{J}$.

c) Calculation of period of oscillations

According to the law of energy conservation,

Hence, the acceleration of the center of mass is given as:

$$\begin{gathered} \frac{dE}{dt}=0 \\ \frac{d}{dt}\left(\frac{1}{2}k{x}^2-\frac{3}{4}M{v}^2\right)=0(frompart(a\left)\right) \\ -\frac{3}{2}M{v}_{cm}{a}_{cm}+kx{v}_{cm}=0 \\ {a}_{cm}=\left(\frac{2k}{3M}\right)x \end{gathered}$$

The period of oscillation for SHM using equation (iv) and the value of acceleration with being L=x is given as:

$T=2\tau \tau \sqrt{\frac{3M}{2k}}$

Therefore, the period of the SHM executes by cylinder’s center of mass is,

$T=2\tau \tau \sqrt{\frac{3M}{2k}}$