Problem 14 - University Physics with Modern Physics

Question

A stone is suspended from the free end of a wire that is wrapped around the outer rim of a pulley, similar to what is shown in Fig. \(10.10 .\) The pulley is a uniform disk with mass 10.0 \(\mathrm{kg}\) and radius 50.0 \(\mathrm{cm}\) and turns on frictionless bearings. You measure that the stone travels 12.6 \(\mathrm{m}\) in the first 3.00 s starting from rest. Find (a) the mass of the stone and (b) the tension in the wire.

Step-by-Step Solution

Answer

The stone's mass is 2 kg and the tension in the wire is 14 N.

Step 1: Understand the Problem

We have a disk-shaped pulley without friction and a stone attached to a wire wrapped around the pulley. The stone starts from rest and travels 12.6 m in 3 seconds. We need to find the mass of the stone and the tension in the wire.

Step 2: Set Up Equations of Motion

Since the stone travels 12.6 m in 3 seconds starting from rest, we use the kinematic equation: \[ s = ut + \frac{1}{2}at^2 \] where \( s = 12.6 \) m, \( u = 0 \) \( m/s \) (initial velocity), and \( t = 3 \) s. Solving for the acceleration \( a \), we have:\[ 12.6 = 0 + \frac{1}{2}a(3)^2 \] \[ 12.6 = \frac{9}{2}a \] \[ a = \frac{12.6 \times 2}{9} = 2.8 \ \mathrm{m/s^2} \]

Step 3: Relate Forces and Torque

For the stone, the force equation is: \[ mg - T = ma \]For the pulley (rotating disk), torque \( \tau \) is related to tension \( T \) by:\[ \tau = I \alpha \] Since \( \alpha \) (angular acceleration) is related to linear acceleration by \( a = \alpha R \), then:\[ I = \frac{1}{2}MR^2 = \frac{1}{2}(10)(0.5)^2 = 1.25 \ \mathrm{kg \, m^2} \]\[ T \cdot R = I \alpha = I \left( \frac{a}{R} \right) = 1.25 \times \frac{2.8}{0.5} \]

Step 4: Solve for Tension and Mass

From torque equation, we have:\[ T \cdot 0.5 = 1.25 \times 5.6 = 7 \] \[ T = \frac{7}{0.5} = 14 \ \mathrm{N} \] Substituting back to the force equation:\[ mg - 14 = ma \] \[ m \times 9.8 - 14 = m \times 2.8 \] \[ m \times (9.8 - 2.8) = 14 \] \[ m \times 7 = 14 \] \[ m = \frac{14}{7} = 2 \ \mathrm{kg} \]

Step 5: Conclusion

The mass of the stone is 2 kg and the tension in the wire is 14 N. We found these by using the linear motion equations, force equilibrium, and rotational dynamics principles.

Key Concepts

Kinematic Equations

Kinematic equations are a set of equations that describe the motion of objects under constant acceleration. These equations allow us to calculate various parameters, such as distance, velocity, and time. In this exercise, since the stone starts from rest and travels a distance in a given time, we can use the kinematic equation:

• \( s = ut + \frac{1}{2}at^2 \)

Here, \( s \) is the distance traveled, \( u \) is the initial velocity (which is 0 in this case), \( a \) is the acceleration, and \( t \) is the time.
By plugging in the values given in the problem—distance \( s = 12.6 \, \mathrm{m} \) and time \( t = 3 \, \mathrm{s} \)—we can solve for the linear acceleration \( a \).
These equations provide a way to understand motion and are crucial when objects are moving in a straight line with a uniform acceleration.

Torque

Torque is pivotal when it comes to rotational motion, similar to how force applies to linear motion. Torque is defined as the rotational equivalent of a force that causes an object to rotate around an axis. It is calculated as:

• \( \tau = r \times F \)

where \( r \) is the distance from the pivot point to the point where the force is applied, and \( F \) is the force applied.
In the problem, the tension in the wire creates a torque on the pulley, forcing it to rotate. This is expressed as:

• \( T \cdot R = I \alpha \)

where \( T \) is the tension in the wire, \( R \) is the pulley's radius, and \( \alpha \) is the angular acceleration. It’s essential to understand torque to analyze and predict the rotational dynamics of systems effectively.

Angular Acceleration

Angular acceleration is the rate of change of angular velocity over time. Just as linear acceleration measures changes in speed in a straight line, angular acceleration does the same for circular motion. It is related to the linear acceleration that we calculated from the kinematic equations by the formula:

• \( a = \alpha R \)

where \( \alpha \) is the angular acceleration and \( R \) is the radius of the circular path.
For the pulley, the angular acceleration can be determined using:

• \( \alpha = \frac{a}{R} \)

This relationship helps understand how forces applied tangentially on a rotating body affect its rotational speed, providing insight into how rotational and linear dynamics are interconnected.

Frictionless Pulley

A frictionless pulley is an idealized concept in physics, used to simplify calculations by assuming there is no friction resisting the rotation of the pulley. In real-life scenarios, pulleys do have friction, but by considering an ideal frictionless scenario, the focus can be placed solely on the forces and torques without additional complexities.
In this exercise, the frictionless pulley ensures that the only forces at play are the force of gravity on the stone and the tension in the wire. This simplification allows us to set up the equations more straightforwardly and directly relate the tension in the wire to the weight of the stone and its acceleration.
Using a frictionless pulley model helps students understand the core principles of rotational dynamics without being sidetracked by the influences of friction.