- #1

brochesspro

- 155

- 22

- Homework Statement
- A solid sphere rolling on a rough horizontal surface with a linear speed ##v_0## collides elastically with a fixed, smooth, vertical wall. Find the speed of the sphere after it has started pure rolling in the backward direction.

- Relevant Equations
- Conservation of angular momentum.

The moment of inertia of a solid sphere of mass ##m## and radius ##R## is ##\frac 2 5mR^2##.

Here is the problem statement along with the figure.

Here, I take the right-ward and anti-clockwise directions to be positive.

After the ball collides with the wall, its angular velocity remains the same and its velocity changes direction while remaining the same in magnitude.

Using the conservation of angular momentum about the bottom-most point, I get:

##-mv_0R -\frac 2 5mR^2\omega_0 = -mv_fR + \frac 2 5 mR^2\omega_f##

##\Rightarrow -mv_0R -\frac 2 5mv_0R = -mv_fR + \frac 2 5 mv_fR##

since ##v_0 = R\omega_0## and ##v_f = R\omega_f##

##\Rightarrow -\frac 7 5 mv_0R = -\frac 3 5 mv_fR##

##\Rightarrow 7v_0 = 3v_f##

##\Rightarrow v_f = \frac 7 3 v_0##

This is clearly false as the sphere slows down due to friction acting in the positive direction. May I know what exactly I have done wrong?

Thanks.

Here, I take the right-ward and anti-clockwise directions to be positive.

After the ball collides with the wall, its angular velocity remains the same and its velocity changes direction while remaining the same in magnitude.

Using the conservation of angular momentum about the bottom-most point, I get:

##-mv_0R -\frac 2 5mR^2\omega_0 = -mv_fR + \frac 2 5 mR^2\omega_f##

##\Rightarrow -mv_0R -\frac 2 5mv_0R = -mv_fR + \frac 2 5 mv_fR##

since ##v_0 = R\omega_0## and ##v_f = R\omega_f##

##\Rightarrow -\frac 7 5 mv_0R = -\frac 3 5 mv_fR##

##\Rightarrow 7v_0 = 3v_f##

##\Rightarrow v_f = \frac 7 3 v_0##

This is clearly false as the sphere slows down due to friction acting in the positive direction. May I know what exactly I have done wrong?

Thanks.