If the angular acceleration of disc just after released assuming that it starts pire rolling and does not slip is ng/717R then find n

To find the value of \( n \) in the expression for angular acceleration of a disc just after it is released, we need to analyze the situation using the principles of rotational dynamics and the conditions for rolling without slipping.

Understanding the Problem

When a disc is released from rest and begins to roll without slipping, it experiences both translational and rotational motion. The angular acceleration (\( \alpha \)) can be related to the linear acceleration (\( a \)) of the center of mass of the disc. The relationship between these two accelerations is given by:

• \( a = R \alpha \)

where \( R \) is the radius of the disc. The disc also experiences a torque due to gravity acting at its center of mass, which leads to angular acceleration.

Applying Newton's Laws

For a disc of mass \( m \) and radius \( R \), the gravitational force acting on it is \( mg \), where \( g \) is the acceleration due to gravity. The torque (\( \tau \)) about the center of mass due to this force can be expressed as:

• \( \tau = R \cdot F = R \cdot mg \)

Using the moment of inertia (\( I \)) for a solid disc, which is given by:

• \( I = \frac{1}{2} m R^2 \)

We can relate torque to angular acceleration using the equation:

• \( \tau = I \alpha \)

Setting Up the Equations

Substituting the expressions for torque and moment of inertia into the equation gives us:

• \( R \cdot mg = \left(\frac{1}{2} m R^2\right) \alpha \)

Now, we can simplify this equation. First, we can cancel \( m \) from both sides (assuming \( m \neq 0 \)):

• \( Rg = \frac{1}{2} R^2 \alpha \)

Next, we can divide both sides by \( R \) (again assuming \( R \neq 0 \)):

• \( g = \frac{1}{2} R \alpha \)

Finding Angular Acceleration

Now, we can solve for \( \alpha \):

• \( \alpha = \frac{2g}{R} \)

According to the problem, the angular acceleration is given as:

• \( \alpha = \frac{ng}{717R} \)

Equating the Two Expressions

Now we can set the two expressions for angular acceleration equal to each other:

• \( \frac{2g}{R} = \frac{ng}{717R} \)

We can cancel \( g \) and \( R \) from both sides (assuming they are not zero):

• \( 2 = \frac{n}{717} \)

Solving for n

To find \( n \), we can multiply both sides by 717:

• \( n = 2 \times 717 \)

Calculating this gives:

• \( n = 1434 \)

Final Result

Thus, the value of \( n \) is \( 1434 \). This means that the angular acceleration of the disc just after it is released, assuming it starts pure rolling and does not slip, is \( \frac{1434g}{717R} \).