A massless string is wrapped around a frictionless pulley of mass $M$. The string is pulled down with a force of 50 N, so that the pulley rotates due to the pull. Consider a point $P$ on the rim of the pulley, which is a solid cylinder. The point has a constant linear (tangential) acceleration component equal to the acceleration of gravity on Earth, which is where this experiment is being held. What is the weight of the cylindrical pulley, in Newtons? 

[i](Proposed by Ahaan Rungta)[/i]

[hide="Note"]
This problem was not fully correct. Within friction, the pulley cannot rotate. So we responded: 
[quote]Excellent observation! This is very true. To submit, I'd say just submit as if it were rotating and ignore friction. In some effects such as these, I'm pretty sure it turns out that friction doesn't change the answer much anyway, but, yes, just submit as if it were rotating and you are just ignoring friction. [/quote]So do this problem imagining that the pulley does rotate somehow. 
[/hide]
1. **Identify the given information and the required quantity:**
   - The force applied on the string: \( F = 50 \, \text{N} \)
   - The tangential acceleration of a point on the rim of the pulley: \( a = g \)
   - The mass of the pulley: \( M \)
   - The radius of the pulley: \( r \)
   - The moment of inertia of a solid cylindrical pulley: \( I = \frac{1}{2} M r^2 \)

2. **Relate the tangential acceleration to the angular acceleration:**
   - The tangential acceleration \( a \) is related to the angular acceleration \( \alpha \) by the equation:
     \[
     a = r \alpha
     \]
   - Given \( a = g \), we can solve for \( \alpha \):
     \[
     \alpha = \frac{a}{r} = \frac{g}{r}
     \]

3. **Apply Newton's second law for rotation:**
   - The torque \( \tau \) exerted by the force \( F \) on the pulley is:
     \[
     \tau = F \cdot r
     \]
   - Using the rotational form of Newton's second law, \( \tau = I \alpha \), we have:
     \[
     F \cdot r = I \alpha
     \]
   - Substitute the moment of inertia \( I = \frac{1}{2} M r^2 \) and the angular acceleration \( \alpha = \frac{g}{r} \):
     \[
     F \cdot r = \left( \frac{1}{2} M r^2 \right) \left( \frac{g}{r} \right)
     \]

4. **Simplify the equation to solve for \( M \):**
   \[
   F \cdot r = \frac{1}{2} M r^2 \cdot \frac{g}{r}
   \]
   \[
   F \cdot r = \frac{1}{2} M r g
   \]
   \[
   50 \cdot r = \frac{1}{2} M r g
   \]
   \[
   50 = \frac{1}{2} M g
   \]
   \[
   M = \frac{100}{g}
   \]

5. **Calculate the weight of the pulley:**
   - The weight \( W \) of the pulley is given by \( W = M g \):
     \[
     W = \left( \frac{100}{g} \right) g = 100 \, \text{N}
     \]

The final answer is \( \boxed{100 \, \text{N}} \).