A disc of mass M and Radius R is rolling with an angular speed `omega` on the horizontal plane. The magnitude of angular momentum of the disc about origin is:

Options

A. `1/2 MR^(2)omega`
B. `MR^(2)omega`
C. `3/2 MR^(2) omega`
D. `2MR^(2) omega`

Explanation

To find the magnitude of the angular momentum of a disc of mass \( M \) and radius \( R \) rolling with an angular speed \( \omega \) about the origin, we can follow these steps:

### Step 1: Understand the Concept of Angular Momentum
Angular momentum \( L \) of a rotating body can be expressed as:
\[ L = I \omega \]
where \( I \) is the moment of inertia of the body and \( \omega \) is the angular speed.

### Step 2: Calculate the Moment of Inertia
For a solid disc about its center of mass, the moment of inertia \( I_{cm} \) is given by:
\[ I_{cm} = \frac{1}{2} M R^2 \]

### Step 3: Apply the Parallel Axis Theorem
Since we need the moment of inertia about the origin (which is a distance \( R \) away from the center of the disc), we will use the Parallel Axis Theorem. The theorem states:
\[ I = I_{cm} + Md^2 \]
where \( d \) is the distance from the center of mass to the new axis (in this case, \( d = R \)). Therefore, we have:
\[ I = \frac{1}{2} M R^2 + M R^2 \]
\[ I = \frac{1}{2} M R^2 + \frac{2}{2} M R^2 = \frac{3}{2} M R^2 \]

### Step 4: Substitute into the Angular Momentum Formula
Now that we have the moment of inertia about the origin, we can substitute it into the angular momentum formula:
\[ L = I \omega = \left(\frac{3}{2} M R^2\right) \omega \]

### Step 5: Final Expression for Angular Momentum
Thus, the magnitude of the angular momentum of the disc about the origin is:
\[ L = \frac{3}{2} M R^2 \omega \]

### Summary
The magnitude of the angular momentum of the disc about the origin is:
\[ L = \frac{3}{2} M R^2 \omega \]

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