Question

A disc is spun with angular velocity $\omega$. What distance should the disc travel so that the velocity at P is instantaneously zero?

• A. R
• B. $\frac{\omega^2R^2}{2g}$
• C. $\sqrt{\frac{\omega R}{g}}$
• D. none

Answer 1

Answer: (B)

$\frac{\omega^2R^2}{2g}$

Answer 2

The problem asks for the distance the disc travels so that the velocity of a point P on its circumference becomes instantaneously zero.

1. Identify the condition for zero velocity at point P: For the velocity of point P to be instantaneously zero, the velocity of the center of mass ($V_c$) must be equal in magnitude and opposite in direction to the rotational velocity of P relative to the center of mass ($V_{P,rot}$). That is, $\vec{V}_c = -\vec{V}_{P,rot}$.

2. Consider the motion: The problem implies the disc is moving under gravity. Let's assume the disc is dropped from rest ($V_{c,0}=0$) and simultaneously given an angular velocity $\omega$. The center of mass of the disc will accelerate downwards due to gravity.

   • Velocity of center of mass: $V_c = gt$ (downwards, taking positive direction downwards).

3. Analyze rotational velocity of P: The magnitude of the rotational velocity of any point on the circumference is $V_{P,rot} = \omega R$. For $\vec{V}_c = -\vec{V}_{P,rot}$ to hold, $\vec{V}_{P,rot}$ must be vertically upwards.

   • This occurs when point P is at the extreme horizontal position (e.g., at the "3 o'clock" or "9 o'clock" position if looking at the disc face-on, with the center at the origin), and the direction of spin $\omega$ is appropriately chosen to give an upward rotational velocity component.
   • At this specific point P, its rotational velocity is purely vertical with magnitude $\omega R$.

4. Equate velocities: For $V_P = 0$, we must have $V_c = V_{P,rot}$.

   • So, $gt = \omega R$.

5. Calculate time and distance:

   • The time at which this condition is met is $t = \frac{\omega R}{g}$.
   • The distance traveled (fallen) by the disc's center of mass is given by $h = V_{c,0}t + \frac{1}{2}gt^2$. Since $V_{c,0}=0$: $h = \frac{1}{2}g \left(\frac{\omega R}{g}\right)^2$ $h = \frac{1}{2}g \frac{\omega^2 R^2}{g^2}$ $h = \frac{\omega^2 R^2}{2g}$