91Ó°ÊÓ

A physics student of mass \(43.0 \mathrm{~kg}\) is standing at the edge of the flat roof of a building, \(12.0 \mathrm{~m}\) above the sidewalk. An unfriendly \(\operatorname{dog}\) is running across the roof toward her. Next to her is a large wheel mounted on a horizontal axle at its center. The wheel, used to lift objects from the ground to the roof, has a light crank attached to it and a light rope wrapped around it; the free end of the rope hangs over the edge of the roof. The student grabs the end of the rope and steps off the roof. If the wheel has radius \(0.300 \mathrm{~m}\) and a moment of inertia of \(9.60 \mathrm{~kg} \cdot \mathrm{m}^{2}\) for rotation about the axle, how long does it take her to reach the sidewalk, and how fast will she be moving just before she lands? Ignore friction in the axle.

Step by step solution

01

Identifying Forces

The force of gravity acting on the student is equal to her mass times the acceleration due to gravity ( \( F_{\text{gravity}} = m \cdot g \), where \( m = 43.0 \, \text{kg} \) and \( g = 9.81 \, \text{m/s}^{2} \)). The force making the wheel turn (torque) is \( r \cdot F \), where \( r = 0.300 \, \text{m} \) is the radius of the wheel and \( F = F_{\text{gravity}} \) is the force due to gravity.

02

Calculating the Acceleration

The angular acceleration of the wheel, \( a \), can be calculated using the equation for torque, \( r \cdot F = I \cdot a \), where \( I = 9.6 \, \text{kg} \cdot \text{m}^{2} \) is the moment of inertia of the wheel. Solving for \( a \) gives \( a = \frac{r \cdot F}{I} \).

03

Calculating the time to Reach the Sidewalk

Using the equations of motion, we can calculate the time it takes to reach the sidewalk. The distance to cover, \( h \), is \( 12.0 \, \text{m} \). The time to reach the ground can be calculated by using \( h = \frac{1}{2} \cdot a \cdot t^{2} \), solving for \( t \) gives \( t = \sqrt{\frac{2h}{a}} \).

04

Calculating the Final Velocity

Her final velocity, \( v \), can be calculated by multiplying the acceleration by the time ( \( v = a \cdot t \) ).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Torque

Imagine trying to push open a heavy door. The force you apply and the point where you push determines how easily the door moves. This principle is similar to the concept of torque in physics. Torque is like a rotational force. It's what causes objects to rotate around an axis.

In our exercise, the student pulls on a rope wrapped around a wheel to rotate it. The torque

• Depends on the force applied, which is the student's weight in this case (\(F = m imes g\)).
• Depends on the distance from the pivot (or axis of rotation) to the point where the force is applied, known as the radius (\(r = 0.300 ext{ m}\)).

The torque formula is expressed as:
\[\text{Torque} = r imes F\]
This rotational force causes the wheel to spin and, importantly, accelerates the student downward as she holds the rope.

Moment of Inertia

When studying rotational motion, the moment of inertia is vital. It is the rotational equivalent of mass in linear motion. Think of it as a measure of how difficult it is to change an object's rotation. For the wheel in our exercise,

• The moment of inertia provides resistance against changes in rotational motion.
• It depends on both the mass distribution of the wheel and its shape.

The formula for moment of inertia can vary based on geometry, but typically,\( I = 9.60 ext{ kg} imes ext{m}^2 \) is used when it's already given for a solid wheel.

In this problem, knowing the moment of inertia helps us calculate the angular acceleration of the wheel. It shows why objects with higher moments of inertia are harder to accelerate rotationally.

Angular Acceleration

When a force causes an object like a wheel to rotate, it also gives rise to angular acceleration. Angular acceleration is how quickly an object's rotation speeds up (or slows down). It is crucial for determining the motion of objects in rotational dynamics.

Let’s break down the calculation:

• The torque applied leads to angular acceleration, expressed as \( \ a = \frac{r \times F}{I} \ \).
• Here, \( r \times F \) represents the torque applied by the student's weight, and \( I \) is the moment of inertia of the wheel.

By calculating this angular acceleration and understanding how the rope interacts with the wheel, we can then predict the linear acceleration of the student downward. This acceleration affects how quickly she reaches the sidewalk and her speed upon landing.
This exercise effectively illustrates the connection between forces in linear motion and those in rotational motion, emphasizing the fundamental role of angular acceleration in analyzing such scenarios.