Problem 25/167:
Given the system shown - all surfaces are frictionless. If m=150 grams is released when it is d=4.0 ft. above the base of M=1650 grams, how long after release will m strike the base of M?
Description of the Drawing:
The drawing shows a system with a block m hanging inside a vertical shaft. The block m is connected by a three pulley system to a larger block M at the bottom where, when m is released, it will pull M horizontally. The block m is initially positioned d=4.0 feet above the base of block M. The entire setup is frictionless.
[latex]To solve this problem, we need to analyze the motion of both masses \( m \) and \( M \) in the pulley system. The key steps involve:
1. **Understanding the pulley system and constraints.**
2. **Applying Newton's second law to each mass.**
3. **Relating the accelerations and distances using the pulley constraints.**
4. **Calculating the time it takes for mass \( m \) to fall the given distance.**
---
### **Step 1: Understand the Pulley System and Constraints**
**Assumptions based on the description:**
- Mass \( m = 150 \) grams (\( 0.150 \) kg) hangs vertically inside a shaft.
- Mass \( M = 1650 \) grams (\( 1.650 \) kg) is on a horizontal frictionless surface.
- The pulley system consists of three pulleys arranged so that when \( m \) moves downward, \( M \) moves horizontally.
- The pulleys provide a mechanical advantage, altering the relationship between the distances and accelerations of \( m \) and \( M \).
**Pulley Constraints:**
- When mass \( m \) moves downward by a distance \( s_m \), mass \( M \) moves horizontally by a distance \( s_M \).
- Based on the pulley system (three pulleys), we can infer that:
\[
s_m = 2 s_M
\]
- This implies that:
\[
v_m = 2 v_M \quad \text{and} \quad a_m = 2 a_M
\]
### **Step 2: Apply Newton's Second Law to Each Mass**
**For Mass \( m \):**
- Forces acting on \( m \):
- Gravity: \( F_g = m g \) (downward)
- Tension: \( T \) (upward)
- Net force:
\[
F_{\text{net}, m} = m g - T = m a_m \tag{1}
\]
**For Mass \( M \):**
- Forces acting on \( M \):
- Tension from two rope segments due to the pulley system: \( 2T \) (horizontal)
- Since the surface is frictionless, there's no frictional force.
- Net force:
\[
F_{\text{net}, M} = 2T = M a_M \tag{2}
\]
### **Step 3: Relate Accelerations and Solve for \( a_m \) and \( T \)**
**From the pulley constraints:**
\[
a_m = 2 a_M \tag{3}
\]
**Solving Equations (1), (2), and (3):**
**Substitute \( a_M = \frac{a_m}{2} \) into Equation (2):**
\[
2T = M \left( \frac{a_m}{2} \right) ==> T = \frac{M a_m}{4} \tag{4}
\]
**Substitute \( T \) from Equation (4) into Equation (1):**
\[
m g - \frac{M a_m}{4} = m a_m
\]
**Solve for \( a_m \):**
\[
m g = m a_m + \frac{M a_m}{4}
\]
\[
m g = a_m \left( m + \frac{M}{4} \right)
\]
\[
a_m = \frac{m g}{m + \frac{M}{4}}
\]
**Plug in the given masses:**
- \( m = 0.150 \) kg
- \( M = 1.650 \) kg
Calculate \( \frac{M}{4} \):
\[
\frac{M}{4} = \frac{1.650\, \text{kg}}{4} = 0.4125\, \text{kg}
\]
Calculate the denominator:
\[
m + \frac{M}{4} = 0.150\, \text{kg} + 0.4125\, \text{kg} = 0.5625\, \text{kg}
\]
Calculate \( m g \):
\[
m g = 0.150\, \text{kg} \times 9.8\, \text{m/s}^2 = 1.47\, \text{N}
\]
Compute \( a_m \):
\[
a_m = \frac{1.47\, \text{N}}{0.5625\, \text{kg}} = 2.6133\, \text{m/s}^2
\]
**Calculate \( a_M \) using Equation (3):**
\[
a_M = \frac{a_m}{2} = \frac{2.6133\, \text{m/s}^2}{2} = 1.3067\, \text{m/s}^2
\]
### **Step 4: Calculate the Time for Mass \( m \) to Fall Distance \( d \)**
**Given:**
- Distance \( d = 4.0\, \text{ft} = 1.2192\, \text{m} \)
- Mass \( m \) starts from rest (\( u = 0 \))
- Acceleration \( a_m = 2.6133\, \text{m/s}^2 \)
**Use the kinematic equation:**
\[
s_m = u t + \frac{1}{2} a_m t^2
\]
Since \( u = 0 \):
\[
s_m = \frac{1}{2} a_m t^2
\]
**But from pulley constraints, \( s_m = 2 s_M \), and we need \( s_m = d = 1.2192\, \text{m} \).**
**Solve for \( t \):**
\[
t = \sqrt{\frac{2 s_m}{a_m}} = \sqrt{\frac{2 \times 1.2192\, \text{m}}{2.6133\, \text{m/s}^2}}
\]
\[
t = \sqrt{\frac{2.4384\, \text{m}}{2.6133\, \text{m/s}^2}} = \sqrt{0.9333\, \text{s}^2}
\]
\[
t \approx 0.966\, \text{s}
\]
---
**Answer:**
Approximately **0.966 seconds** after release, mass \( m \) will strike the base of mass \( M \).
---
**Note:** This solution assumes a specific pulley arrangement based on the description provided. If the actual pulley system differs, the relationships between distances and accelerations might change, affecting the final result.
Problem 50/167:
A 100-ft. rope weighing 0.2 lb./ft. is stretched out on the horizontal deck of a ship. A 40 lb. anchor is attached to one end and is released overboard. Assuming that the rope is perfectly flexible, that the speed of the rope and anchor is initially zero, and that the coefficient of friction between rope and deck is 0.4, how fast is the anchor falling when it is 50 feet below the deck?
[latex]To solve this problem, we'll apply the **work-energy principle**, which states that the work done by all forces acting on an object equals the change in its kinetic energy. This method is suitable because friction and variable mass are involved, and the system's energy changes as the rope and anchor move.
**Given:**
- Total length of rope, \( L = 100 \) ft
- Weight per unit length of rope, \( \lambda = 0.2 \) lb/ft
- Total weight of rope: \( W_{\text{rope}} = \lambda L = 20 \) lb
- Weight of anchor, \( W_{\text{anchor}} = 40 \) lb
- Coefficient of friction between rope and deck, \( \mu = 0.4 \)
- Distance fallen by anchor, \( s = 50 \) ft
- Gravitational acceleration, \( g = 32.174 \) ft/s²
---
### **Step 1: Calculate the Total Potential Energy Lost**
**Potential energy lost by the anchor:**
\[
\Delta PE_{\text{anchor}} = W_{\text{anchor}} \times s = 40\, \text{lb} \times 50\, \text{ft} = 2000\, \text{ft}\cdot\text{lb}
\]
**Potential energy lost by the rope:**
The rope hanging overboard has a length \( s \) and weight \( \lambda s \). Its center of mass descends from the deck to \( \frac{s}{2} \) below the deck. Thus:
\[
\Delta PE_{\text{rope}} = \lambda s \times \left( \frac{s}{2} \right) = \frac{\lambda s^2}{2} = \frac{0.2\, \text{lb/ft} \times (50\, \text{ft})^2}{2} = 250\, \text{ft}\cdot\text{lb}
\]
**Total potential energy lost:**
\[
\Delta PE_{\text{total}} = \Delta PE_{\text{anchor}} + \Delta PE_{\text{rope}} = 2000\, \text{ft}\cdot\text{lb} + 250\, \text{ft}\cdot\text{lb} = 2250\, \text{ft}\cdot\text{lb}
\]
---
### **Step 2: Calculate the Work Done Against Friction**
The friction force varies as the rope leaves the deck:
\[
F_{\text{friction}}(s) = \mu \lambda (L - s)
\]
**Total work done against friction:**
\[
\begin{align*}
W_{\text{friction}} &= \int_{0}^{50} F_{\text{friction}}(s) \, ds = \int_{0}^{50} \mu \lambda (L - s) \, ds \\
&= \mu \lambda \left[ Ls - \frac{s^2}{2} \right]_0^{50} \\
&= 0.4 \times 0.2\, \text{lb/ft} \left[ 100\, \text{ft} \times 50\, \text{ft} - \frac{(50\, \text{ft})^2}{2} \right] \\
&= 0.08 \times \left( 5000\, \text{ft}^2 - 1250\, \text{ft}^2 \right) = 0.08 \times 3750\, \text{ft}^2 = 300\, \text{ft}\cdot\text{lb}
\end{align*}
\]
---
### **Step 3: Apply the Work-Energy Principle**
**Net work done:**
\[
\text{Net Work} = \Delta PE_{\text{total}} - W_{\text{friction}} = 2250\, \text{ft}\cdot\text{lb} - 300\, \text{ft}\cdot\text{lb} = 1950\, \text{ft}\cdot\text{lb}
\]
This net work converts into kinetic energy:
\[
KE_{\text{final}} = \frac{1}{2} m_{\text{total}} v^2 = 1950\, \text{ft}\cdot\text{lb}
\]
---
### **Step 4: Calculate the Total Mass in Motion**
**Mass of the anchor:**
\[
m_{\text{anchor}} = \frac{W_{\text{anchor}}}{g} = \frac{40\, \text{lb}}{32.174\, \text{ft/s}^2} \approx 1.243\, \text{slugs}
\]
**Mass of the hanging rope:**
\[
m_{\text{rope}} = \frac{\lambda s}{g} = \frac{0.2\, \text{lb/ft} \times 50\, \text{ft}}{32.174\, \text{ft/s}^2} \approx 0.311\, \text{slugs}
\]
**Total moving mass:**
\[
m_{\text{total}} = m_{\text{anchor}} + m_{\text{rope}} \approx 1.243\, \text{slugs} + 0.311\, \text{slugs} = 1.554\, \text{slugs}
\]
---
### **Step 5: Solve for the Final Velocity**
Rearrange the kinetic energy equation:
\[
v^2 = \frac{2 KE_{\text{final}}}{m_{\text{total}}} = \frac{2 \times 1950\, \text{ft}\cdot\text{lb}}{1.554\, \text{slugs}} \approx \frac{3900\, \text{ft}\cdot\text{lb}}{1.554\, \text{slugs}}
\]
Since \( 1\, \text{lb} = 1\, \text{slug} \times \text{ft/s}^2 \):
\[
v^2 \approx \frac{3900\, \text{lb}\cdot\text{ft}}{1.554\, \text{slugs}} = \frac{3900\, \text{slug}\cdot\text{ft/s}^2 \cdot \text{ft}}{1.554\, \text{slugs}} = \frac{3900\, \text{ft}^2/\text{s}^2}{1.554}
\]
\[
v^2 \approx 2508.367\, \text{ft}^2/\text{s}^2 ==> v \approx \sqrt{2508.367}\, \text{ft/s} \approx 50.08\, \text{ft/s}
\]
---
**Answer:**
The anchor is falling at approximately **50 feet per second** when it is 50 feet below the deck.
---
**Final Answer:**
Approximately **50 feet per second**.
Problem 101/167:
[latex]To determine the torsion constant (\( \kappa \)) of the fiber, we'll use the torsional oscillation formula and calculate the moment of inertia (\( I \)) of the cylindrical rod.
### **Given:**
- Mass of the rod, \( m = 200 \) g \( = 0.200 \) kg
- Length of the rod, \( L = 10 \) cm \( = 0.10 \) m
- Diameter of the rod, \( d = 1.0 \) cm \( = 0.010 \) m
- Radius of the rod, \( R = \frac{d}{2} = 0.005 \) m
- Period of oscillation, \( T = 3.0 \) s
### **Step 1: Calculate the Moment of Inertia (\( I \))**
Since the rod is not "slim," we use the moment of inertia for a solid cylinder rotating about an axis perpendicular to its length and passing through its center:
\[
I = \frac{1}{12} m (3R^2 + L^2)
\]
Compute \( R^2 \) and \( L^2 \):
\[
R^2 = (0.005\, \text{m})^2 = 0.000025\, \text{m}^2
\]
\[
L^2 = (0.10\, \text{m})^2 = 0.01\, \text{m}^2
\]
Now calculate \( I \):
\[
I = \frac{1}{12} (0.200\, \text{kg}) [3 \times 0.000025\, \text{m}^2 + 0.01\, \text{m}^2]
\]
\[
I = \frac{0.200\, \text{kg}}{12} (0.000075\, \text{m}^2 + 0.01\, \text{m}^2)
\]
\[
I = \frac{0.200\, \text{kg}}{12} \times 0.010075\, \text{m}^2
\]
\[
I = \frac{0.002015\, \text{kg}\cdot\text{m}^2}{12}
\]
\[
I = 0.0001679167\, \text{kg}\cdot\text{m}^2
\]
### **Step 2: Calculate the Torsion Constant (\( \kappa \))**
The period \( T \) of torsional oscillation is related to \( I \) and \( \kappa \) by:
\[
T = 2\pi \sqrt{\frac{I}{\kappa}}
\]
Rewriting for \( \kappa \):
\[
\kappa = \frac{4\pi^2 I}{T^2}
\]
Compute \( T^2 \) and \( 4\pi^2 \):
\[
T^2 = (3.0\, \text{s})^2 = 9.0\, \text{s}^2
\]
\[
4\pi^2 = 4 \times (\pi)^2 \approx 4 \times 9.8696 = 39.4784
\]
Now calculate \( \kappa \):
\[
\kappa = \frac{39.4784 \times 0.0001679167\, \text{kg}\cdot\text{m}^2}{9.0\, \text{s}^2}
\]
\[
\kappa = \frac{0.00663212\, \text{kg}\cdot\text{m}^2}{9.0\, \text{s}^2}
\]
\[
\kappa = 0.0007369\, \text{N}\cdot\text{m}/\text{rad}
\]
### **Answer:**
**The torsion constant of the fiber is approximately \( \kappa = 7.37 \times 10^{-4} \) N·m/rad.**
Let's analyze each value:
- From the answer above: 7.37 × 10^-4 N·m/rad
- The test answer: 7.4 x 10^3 dynes cm·rad^-1
To make these comparable, we'll convert the second value to N·m/rad:
1 dyne = 10^-5 N
1 cm = 10^-2 m
7.4 x 10^3 dynes cm·rad^-1
= 7.4 x 10^3 × (10^-5 N) × (10^-2 m) × rad^-1
= 7.4 x 10^-4 N·m/rad
Now we can compare:
7.37 × 10^-4 N·m/rad (first value)
7.4 × 10^-4 N·m/rad (converted second value)
These values are extremely close, differing only in the second decimal place. The slight difference could be due to rounding in the calculations or slight variations in the problem-solving approach. For practical purposes, these values can be considered essentially equivalent, representing the same torsion constant of the fiber in the given problem.
Problem 160/167:
N moles of ideal gas are undergoing compression in a heat-insulated cylinder fitted with a piston of area A. The piston moves with a speed v.
(a) If the cylinder is perfectly insulated except for a small heat leak through which heat can flow, how much heat per second must flow through the leak for the temperature to remain constant at a time when the pressure is P?
(b) If the heat leak is suddenly stopped when the pressure reaches P, what will be the rate of rise of temperature, dT/dt, immediately thereafter?
[latex]To solve this problem, we'll apply the principles of thermodynamics and mechanics to determine:
1. **The rate of heat flow required to keep the gas temperature constant during compression.**
2. **The rate of temperature increase immediately after the heat leak is stopped.**
---
### **Given:**
- Number of moles of gas: \( N \)
- Piston area: \( A \)
- Piston speed: \( v \)
- Gas pressure at the given instant: \( P \)
- Ideal gas behavior is assumed.
---
## **Part (a): Heat Flow Required for Constant Temperature**
**Objective:** Find the rate of heat flow (\( \dot{Q} \)) that must pass through the leak to maintain a constant temperature during compression.
### **Solution Steps:**
1. **Determine the Rate of Work Done on the Gas:**
The piston compresses the gas, doing work on it. The rate at which work is done (\( \dot{W} \)) is:
\[
\dot{W} = \text{Force} \times \text{Velocity} = P \times A \times v
\]
- \( P \): Pressure exerted on the piston.
- \( A \): Area of the piston.
- \( v \): Speed of the piston.
2. **Apply the First Law of Thermodynamics:**
For an isothermal (constant temperature) process, the internal energy (\( U \)) of an ideal gas remains constant because it depends only on temperature:
\[
\Delta U = 0 = Q - W
\]
Therefore, the heat flow into the system (\( Q \)) equals the work done by the gas (\( W \)):
\[
Q = W
\]
3. **Determine the Rate of Heat Flow (\( \dot{Q} \)):**
Since \( Q = W \), the rate at which heat must flow out of the gas to keep the temperature constant is:
\[
\dot{Q} = \dot{W} = P A v
\]
**Answer for Part (a):**
The heat that must flow out per second to keep the temperature constant is:
\[
\boxed{\dot{Q} = P A v}
\]
---
## **Part (b): Rate of Temperature Rise After the Heat Leak is Stopped**
**Objective:** Find the rate of temperature increase (\( \frac{dT}{dt} \)) immediately after the heat leak is stopped.
### **Solution Steps:**
1. **Apply the First Law of Thermodynamics Without Heat Exchange:**
After the heat leak is stopped, the process becomes adiabatic (no heat transfer):
\[
\Delta U = Q - W ==> \Delta U = -W \quad (\text{since } Q = 0)
\]
2. **Express Work Done in Terms of Pressure and Volume Change:**
The work done by the gas during a small compression (\( \delta W \)) is:
\[
\delta W = P \, dV
\]
For compression, \( dV \) is negative (volume decreases).
3. **Relate Change in Internal Energy to Temperature Change:**
For an ideal gas, the change in internal energy (\( \Delta U \)) is:
\[
\Delta U = n C_v \, dT
\]
- \( n \): Number of moles.
- \( C_v \): Molar heat capacity at constant volume.
4. **Set Up the Differential Equation:**
Substituting the expressions for \( \Delta U \) and \( \delta W \):
\[
n C_v \, \frac{dT}{dt} = -P \, \frac{dV}{dt}
\]
5. **Relate the Rate of Volume Change to Piston Speed:**
The rate of volume change (\( \frac{dV}{dt} \)) due to the piston moving inward is:
\[
\frac{dV}{dt} = -A v
\]
The negative sign indicates that the volume is decreasing.
6. **Substitute \( \frac{dV}{dt} \) into the Differential Equation:**
\[
n C_v \, \frac{dT}{dt} = -P \times (-A v) ==> n C_v \, \frac{dT}{dt} = P A v
\]
7. **Solve for the Rate of Temperature Increase (\( \frac{dT}{dt} \)):**
\[
\frac{dT}{dt} = \frac{P A v}{n C_v}
\]
**Answer for Part (b):**
Immediately after the heat leak is stopped, the rate of temperature rise is:
\[
\boxed{\frac{dT}{dt} = \frac{P A v}{n C_v}}
\]
---
**Note:**
- \( C_v \) is the molar heat capacity at constant volume, which for a monatomic ideal gas is \( \frac{3}{2} R \) and for a diatomic gas is \( \frac{5}{2} R \), where \( R \) is the ideal gas constant.
- The positive sign in \( \frac{dT}{dt} \) indicates that the temperature increases during adiabatic compression.
Does the answer for b) equate to: 2 PAv/3 NR?
[latex]Yes, you are correct. The expression for the rate of temperature increase in part (b) does equate to:
\[
\frac{dT}{dt} = \frac{2 P A v}{3 N R}
\]
### **Explanation:**
In part (b), we derived:
\[
\frac{dT}{dt} = \frac{P A v}{N C_v}
\]
For an **ideal monatomic gas**, the molar heat capacity at constant volume (\( C_v \)) is:
\[
C_v = \frac{3}{2} R
\]
Substituting \( C_v = \frac{3}{2} R \) into the expression:
\[
\frac{dT}{dt} = \frac{P A v}{N \left( \dfrac{3}{2} R \right)}
\]
Simplify the denominator:
\[
\frac{dT}{dt} = \frac{P A v}{N \times \dfrac{3}{2} R} = \frac{P A v}{\dfrac{3 N R}{2}}
\]
Invert the fraction in the denominator:
\[
\frac{dT}{dt} = P A v \times \frac{2}{3 N R}
\]
Simplify:
\[
\frac{dT}{dt} = \frac{2 P A v}{3 N R}
\]
---
**Conclusion:**
- **Yes**, the rate of temperature increase immediately after the heat leak is stopped is:
\[
\boxed{\frac{dT}{dt} = \dfrac{2 P A v}{3 N R}}
\]
- This result assumes the gas is **monatomic** and ideal.
---
**Note:** If the gas is **diatomic** or has a different molar heat capacity, the value of \( C_v \) would change accordingly:
- For a diatomic ideal gas:
\[
C_v = \frac{5}{2} R
\]
The expression would then be:
\[
\frac{dT}{dt} = \frac{P A v}{N C_v} = \frac{P A v}{N \left( \dfrac{5}{2} R \right)} = \frac{2 P A v}{5 N R}
\]
---
**Summary:**
- **Monatomic Ideal Gas:**
\[
\frac{dT}{dt} = \frac{2 P A v}{3 N R}
\]
- **Diatomic Ideal Gas:**
\[
\frac{dT}{dt} = \frac{2 P A v}{5 N R}
\]
So, your expression is correct for a monatomic ideal gas undergoing adiabatic compression after the heat leak is stopped.