Question

a solid cylinder of mass 3 kg is rolling on a horizontal surface with velocity 4 m/s.It collides with a horizontal spring of force constant 200 N/m.
what will be the maximum compression produced in spring?

Answer 1

Answer on Question#39598 – Physics – Mechanics

A solid cylinder of mass $3\,\mathrm{kg}$ is rolling on a horizontal surface with velocity $4\,\mathrm{m/s}$. It collides with a horizontal spring of force constant $200\,\mathrm{N/m}$. What will be the maximum compression produced in spring?

Solution:

Here we should apply conservation of mechanical energy.

Kinetic energy of rolling body initially is equal to potential energy of the spring:

$$E_k = E_p \quad (1)$$

Potential energy of the spring ($\Delta x$-maximum compression of the spring):

$$E_p = \frac{k \Delta x^2}{2}$$

Kinetic energy of rolling body = Translational KE + Rotational KE:

$$E_k = \frac{m v^2}{2} + \frac{J \omega^2}{2} \quad (2)$$

Moment of inertia of solid cylinder:

$$J = \frac{m R^2}{2} \quad (3)$$

Angular velocity of the cylinder:

$$\omega = \frac{v}{R} \quad (4)$$

(4) and (3) in (2):

$$E_k = \frac{m v^2}{2} + \frac{m R^2}{2} \frac{\left(\frac{v}{R}\right)^2}{2} = \frac{m v^2}{2} + \frac{m v^2}{4} = \frac{3 m v^2}{4} \quad (5)$$

(5) to (1)

$$\frac{k \Delta x^2}{2} = \frac{3 m v^2}{4}$$ $$\Delta x = \sqrt{\frac{3 m v^2}{2 k}} = \sqrt{\frac{3 \cdot 3 k g \cdot \left(4 \frac{m}{s}\right)^2}{2 \cdot 200 \frac{N}{m}}} = 0.6\,\mathrm{m}$$

Answer: the maximum compression produced in spring will be $0.6\,\mathrm{m}$.