91Ó°ÊÓ

In everyday life, we can think of mass moment of inertia as the "rotational mass" of an object. It's similar to how mass works in linear motion. The mass moment of inertia, denoted by the symbol \( I \), is vital in understanding how an object will behave when it is subjected to rotational forces.
The formula to determine it is \( I = mk^2 \), where \( m \) is the mass and \( k \) is the radius of gyration. The radius of gyration is a bit like the average distance of the mass from the axis of rotation. It essentially tells us how far the mass might be "spread out" from the central point in terms of rotation.
With our exercise here, the mass moment of inertia is calculated as \( I = 6 \, \text{kg} \times (0.09 \, \text{m})^2 = 0.0486 \, \text{kg} \cdot \text{m}^2 \). So the disk's resistance to rotational change, given its mass distribution, is represented by this value. This concept is a cornerstone in dynamics when examining rotating objects.

When people think of Newton's Second Law, they often think about \( F = ma \), but did you know it has a rotational counterpart? This law helps us understand how forces cause changes in rotational motion. When a force causes an object to rotate, we talk about torque, \( \tau \), rather than force. Torque is essentially how much "twist" a force provides.
Newton's Second Law for rotation can be expressed with the formula \( \tau = I\alpha \), where \( I \) is the mass moment of inertia and \( \alpha \) is the angular acceleration. In our problem, the torque exerted by the force \( P \) is calculated as \( \tau = P \cdot r_d = 20 \, \text{N} \times 0.06 \, \text{m} = 1.2 \, \text{N} \cdot \text{m} \). With this torque value, you use \( \tau = I\alpha \) to find the angular acceleration \( \alpha \).
Solve for \( \alpha \), which is \( \alpha = \frac{1.2}{0.0486} \approx 24.69 \text{rad/s}^2 \), letting us know how quickly the rotation speed changes.

Static friction is what prevents objects from sliding off surfaces. It's the force that needs to be overcome for motion to start. It plays a crucial role in rolling without slipping, as in the case of the disk.
The static friction force, \( f_s \), is given by \( f_s = \mu_s N \), where \( \mu_s \) is the static friction coefficient and \( N \) is the normal force. Here, \( N \) is equal to the force due to gravity on the disk, \( mg \), which is \( 6 \, \text{kg} \times 9.81 \, \text{m/s}^2 \), simplifying to a straightforward calculation.
To find the minimum static friction coefficient, \( \mu_s \), needed to prevent slipping, we use \( ma = \mu_s mg \). Solving for \( \mu_s \), substituting \( a = 1.481 \, \text{m/s}^2 \), gives \( \mu_s = \frac{1.481}{9.81} \approx 0.151 \). This value tells us the required stickiness between the disk and the surface to keep it rolling smoothly.