A wheel of radius \(24.7 \mathrm{~cm}\), moving initially at \(43.3 \mathrm{~m} / \mathrm{s}\), rolls to a stop in \(225 \mathrm{~m} .\) Calculate \((a)\) its linear acceleration and \((b)\) its angular acceleration. ( \(c\) ) The wheel's rotational inertia is \(0.155 \mathrm{~kg} \cdot \mathrm{m}^{2} .\) Calculate the torque exerted by rolling friction on the wheel.

Linear acceleration is a measure of how quickly an object's velocity changes in a straight line. It is a vector quantity, meaning it has both magnitude and direction. In the context of our exercise, linear acceleration (denoted as 'a') determines the rate at which the wheel slows down to a stop.

Understanding linear acceleration is crucial because it tells us how forcefully the wheel is decelerating. To find the linear acceleration, we can rearrange the kinematics equation, which in this case is \(v_f^2 = v_i^2 + 2ad\). Solving for 'a' gives us the formula \(a = (v_f^2 - v_i^2) / (2d)\), where \(v_f\) is the final velocity, \(v_i\) is the initial velocity, and 'd' is the displacement.

Angular acceleration (denoted by the Greek letter 'alpha') is the rate of change of angular velocity. In everyday terms, it's how quickly a rotating object speeds up or slows down.

Because the wheel in our exercise is rolling without slipping, there's a direct relationship between the linear and angular accelerations. The equation \(a = r\alpha\) links the linear acceleration 'a' with the angular acceleration 'alpha', where 'r' is the radius of the wheel. By calculating 'a' and knowing the value of 'r', we can easily find 'alpha'. This illustrates how rotational motion can be analogous to linear motion, with angular acceleration corresponding to linear acceleration.

Torque is a measure of how much force acting on an object causes that object to rotate. The equation to calculate torque (denoted by the Greek letter 'tau') is \(\tau = I\alpha\), where 'I' is the rotational inertia and 'alpha' is the angular acceleration.

Understanding torque is vital because it quantifies the rotational effect of forces. In the case of our rolling wheel, the torque exerted by rolling friction causes the wheel to slow down. After finding 'alpha' from the previous step, we use the given value of 'I' to solve for 'tau', giving us insight into the strength of the forces resisting the wheel's motion due to friction.

Rotational inertia (also known as the moment of inertia) represents how difficult it is to change the rotational state of an object. It depends on the mass distribution of the object with respect to its axis of rotation. The larger the rotational inertia, the harder it is to get the object to spin or to stop spinning.

In the provided exercise, the rotational inertia 'I' is given as a constant value. When calculating torque, this value is crucial as it influences the rotational motion directly. If the wheel had more mass at its rim, for example, the rotational inertia would be greater, leading to a need for more torque to achieve the same angular acceleration.

Rolling friction is the force that resists the motion of a rolling object across a surface. It is generally much smaller than sliding friction. For an object rolling without slipping, like our wheel, rolling friction is what slows the roll to a stop.

In our exercise, we consider the rolling friction when calculating the torque because it's the rolling friction that applies the force at the edge of the wheel, resulting in a torque that decelerates the wheel. The torque due to rolling friction is an essential factor in determining how quickly the wheel stops.

Kinematics equations describe the motion of objects without considering the forces causing the motion. These equations relate displacement, initial velocity, final velocity, acceleration, and time.

In the step-by-step solution, we applied the kinematics equation \(v_f^2 = v_i^2 + 2ad\) to find the linear acceleration of the wheel. This equation is one among a set that are useful tools for solving a wide range of motion problems, be they linear or rotational, as long as we can establish a relationship between linear and angular quantities.