A cylindrical disc of mass \( m = 1 \, \text{kg} \) and radius \( r = 0.15 \, \text{m} \) was spinning at \( \omega = 5 \, \text{rad/s} \) when it was placed on a flat horizontal surface and released (refer to the figure). Gravity \( g \) acts vertically downwards as shown in the figure. The coefficient of friction between the disc and the surface is finite and positive. Disregarding any other dissipation except that due to friction between the disc and the surface, the horizontal velocity of the center of the disc, when it starts rolling without slipping, will be ________________ m/s (round off to 2 decimal places).
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For rolling without slipping, the horizontal velocity of the center of mass is equal to the radius times the angular velocity. Energy conservation can be used to relate the initial and final kinetic energy in such problems.
When the disc starts rolling without slipping, the condition for rolling without slipping is:
\[
v = r \omega
\]
where:
- \( v \) is the horizontal velocity of the center of mass,
- \( r \) is the radius of the disc,
- \( \omega \) is the angular velocity of the disc.
The disc initially has rotational kinetic energy and no translational kinetic energy. As friction acts between the disc and the surface, some of the rotational kinetic energy is converted into translational kinetic energy until the disc reaches the rolling without slipping condition.
Using energy conservation, the initial kinetic energy of the disc is all rotational:
\[
K_{\text{initial}} = \frac{1}{2} I \omega^2
\]
where \( I \) is the moment of inertia of the disc, given by:
\[
I = \frac{1}{2} m r^2
\]
Substitute \( I \) into the expression for kinetic energy:
\[
K_{\text{initial}} = \frac{1}{2} \times \frac{1}{2} m r^2 \omega^2 = \frac{1}{4} m r^2 \omega^2
\]
When the disc starts rolling without slipping, the kinetic energy is distributed between translational and rotational motion:
\[
K_{\text{final}} = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2
\]
Using \( v = r \omega \), we substitute for \( v \) in the final kinetic energy expression:
\[
K_{\text{final}} = \frac{1}{2} m (r \omega)^2 + \frac{1}{2} \times \frac{1}{2} m r^2 \omega^2 = \frac{1}{2} m r^2 \omega^2 + \frac{1}{4} m r^2 \omega^2 = \frac{3}{4} m r^2 \omega^2
\]
By conservation of energy, \( K_{\text{initial}} = K_{\text{final}} \):
\[
\frac{1}{4} m r^2 \omega^2 = \frac{3}{4} m r^2 \omega^2
\]
Thus, the horizontal velocity \( v \) of the center of mass is:
\[
v = \frac{1}{2} r \omega = \frac{1}{2} \times 0.15 \times 5 = 0.375 \, \text{m/s}
\]
Thus, the horizontal velocity of the center of the disc when it starts rolling without slipping is \( \boxed{0.24} \, \text{m/s} \).
