Question

A solid cylinder of mass 3kg is rolling on a horizontal surface with velocity $4\,m/s$. It collides with a horizontal spring of force constant $200\,N/m$. The maximum compression produced in the spring will be:
A. $0.7m$
B. $0.2m$
C. $0.5m$
D. $0.6m$

Answer 1

Hint: In this solution, we will first calculate the linear kinetic energy of the horizontal cylinder and then calculate the rotational kinetic energy. Then we will convert the angular velocity in terms of linear velocity and radius, and apply the law of conservation of energy on the system of spring and solid cylinder.

Complete step by step answer:
We know that the potential energy of a spring is given as follows:
\[(\dfrac{1}{2})k{x^2} = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}I{\omega ^2}\]
\[U = (\dfrac{1}{2})k{x^2}\]
Here, U is the potential energy of the spring; k is the spring constant of the spring and x is the displacement of the spring.
The kinetic energy of the solid cylinder is divided into two components, which are transversal and rotational. The kinetic energy of the solid cylinder is due to its velocity, rotational kinetic energy is due to angular velocity of the cylinder. It is given as follows:
\[K = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}I{\omega ^2}\]
Here, K is the total kinetic energy, m is the mass of the solid cylinder, v is the velocity of the cylinder, I is the moment of inertia, andis the angular velocity of the solid cylinder. Now, as we know that energy is not destroyed and not conserved, we apply the law of conservation of energy as follows:
\[\dfrac{1}{2}k{x^2} = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}I{\omega ^2}\]
And here,\[I = \dfrac{1}{2}M{R^2}\]and\[\omega = {(\dfrac{v}{R})^2}\]
Thus,
\[
\dfrac{1}{2}k{x^2} = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}(\dfrac{1}{2}M{R^2}){(\dfrac{v}{{{R^{}}}})^2} \\
\Rightarrow \dfrac{1}{2}k{x^2} = \dfrac{3}{4}m{v^2} \\
\Rightarrow \dfrac{1}{2}(200){x^2} = \dfrac{3}{4}(3)(16) \\
\therefore {x^{}} = 0.6m
\]
Thus the spring is compressed up to 0.6m.

Hence option D is the correct answer.

Note:Here, we consider that there is enough friction on the horizontal surface so that the cylinder rolls on the surface and does not slide in its movement. Additionally, we must not forget to take into account the rotational kinetic energy as well as the linear kinetic energy while applying the law of conservation of energy.