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Magazine Chapter 7: Problem 73
A cyliinder of mass \(M\) and radius \(R\) starts falling frecly undcr gravity at \(t=0\) as shown in figure. The tension in cach string at any given time \(t\) is \(T\). The instantancous power developed by the gravitational forec at time \(/\) is \(P\). The linear acceleration of the cylinder is \(a\). Which of the following are correct? (a) \(T-(m g / 6)\) (b) \(a-(2 g / 3)\) (c) \(P-\frac{2}{3} m g^{2} t\) (d) None
Short Answer
Expert verified
Options (b) \(a - \frac{2g}{3}\) and (c) \(P - \frac{2}{3} mg^{2} t\) are correct.
Step by step solution
01
Analyzing forces on the cylinder
The gravitational force acting on the cylinder is given by the weight of the cylinder, which is \( F = Mg \). This force acts downward. The tension \( T \) in each string acts upward, opposing the motion. Since the cylinder is falling freely under gravity, the net force acting on it can be used to find the linear acceleration using Newton's second law.
02
Applying Newton's second law
According to Newton's second law, the net force acting on the cylinder is \( F_{ ext{net}} = Mg - T \). This net force is also equal to the mass times the linear acceleration of the cylinder, or \( Ma \). Therefore, we can write the equation as:\[ Mg - T = Ma \]Rearrange this equation to express tension \( T \) in terms of acceleration \( a \):\[ T = Mg - Ma \]
03
Analyzing rotational motion
Because the cylinder is falling with an acceleration \( a \), it also has an angular acceleration \( \alpha \). The relationship between linear acceleration \( a \) and angular acceleration \( \alpha \) is given by \( a = R\alpha \). The torque \( \tau \) caused by the tension \( T \) is \( \tau = TR \), and this is equal to the moment of inertia \( I \) times angular acceleration \( \alpha \). For a solid cylinder, \( I = \frac{1}{2}MR^2 \).Equating these gives:\[ TR = \frac{1}{2}MR^2 \alpha \]Substitute \( \alpha = \frac{a}{R} \):\[ TR = \frac{1}{2}MR^2 \frac{a}{R} \]Simplify to:\[ T = \frac{1}{2}Ma \]
04
Relating acceleration
From \( T = Mg - Ma \) and \( T = \frac{1}{2}Ma \), set the equations equal and solve for \( a \) as follows:\[ Mg - Ma = \frac{1}{2}Ma \]Combine terms:\[ Mg = \frac{3}{2}Ma \]Solve for \( a \):\[ a = \frac{2}{3}g \]
05
Calculating power
The instantaneous power \( P \) developed by the gravitational force is the force times the velocity, or \( P = Fv \), where velocity \( v = at = \frac{2}{3}gt \) (using \( a = \frac{2}{3}g \) from the previous step). Therefore:\[ P = Mg \left( \frac{2}{3}gt \right) = \frac{2}{3}Mg^2t \]
06
Evaluating provided options
Let us evaluate the given options with the results derived:- (a) "\( T - \frac{mg}{6} \)": From \( T = \frac{1}{2}Ma \) and substituting \( a = \frac{2}{3}g \), we get \( T = \frac{1}{2}M \frac{2}{3}g = \frac{Mg}{3} \). This does not simplify to \( \frac{mg}{6} \).- (b) "\( a - \frac{2g}{3} \)": We derived \( a = \frac{2}{3}g \), so this is correct.- (c) "\( P - \frac{2}{3}mg^2t \)": We calculated \( P = \frac{2}{3}Mg^2t \) which matches, making this correct.- (d) None: Since (b) and (c) are correct, this option is incorrect.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Newton's second law
Newton's second law is a fundamental principle in physics and helps us understand motion. It states that the force acting on an object is equal to the mass of that object multiplied by its acceleration:
\[ F = ma \]
In our case, the cylinder falls freely under gravity.
\[ F_{\text{net}} = Ma \]
This relationship allows us to fin the tension \( T \) and linear acceleration \( a \). Solving for \( T \), we find that:
\[ T = Mg - Ma \]
Through this understanding, we can explore how forces and motion interplay in rotational systems.
\[ F = ma \]
In our case, the cylinder falls freely under gravity.
- The weight of the cylinder (gravitational force) is \( Mg \).
- The tension \( T \) in the strings opposes this force.
- The net force affecting the cylinder is \( F_{\text{net}} = Mg - T \).
\[ F_{\text{net}} = Ma \]
This relationship allows us to fin the tension \( T \) and linear acceleration \( a \). Solving for \( T \), we find that:
\[ T = Mg - Ma \]
Through this understanding, we can explore how forces and motion interplay in rotational systems.
torque and angular acceleration
In rotational dynamics, torque and angular acceleration are equivalent concepts to force and linear acceleration in linear dynamics. When the cylinder falls:
\[ TR = I\alpha \]
Substituting \( \alpha = \frac{a}{R} \) gives:
\[ TR = \frac{1}{2}MR^2 \frac{a}{R} \]
From simplification, we get the tension in terms of acceleration:
\[ T = \frac{1}{2}Ma \]
This shows us how linear forces relate to angular changes through torque and angular acceleration.
- There is both linear motion and rotational motion.
- The linear acceleration \( a \) translates to angular acceleration \( \alpha \) via \( a = R\alpha \).
- The torque \( \tau \) caused by tension is \( \tau = TR \).
\[ TR = I\alpha \]
Substituting \( \alpha = \frac{a}{R} \) gives:
\[ TR = \frac{1}{2}MR^2 \frac{a}{R} \]
From simplification, we get the tension in terms of acceleration:
\[ T = \frac{1}{2}Ma \]
This shows us how linear forces relate to angular changes through torque and angular acceleration.
moment of inertia
Moment of inertia is a measure of an object's resistance to changes in its rotation. Just like mass is a measure of resistance to changes in linear motion:
\[ TR = I \alpha \]
The solution of this expression highlights how the cylinder's distribution of mass affects its rotational movement. Moments of inertia are unique for different shapes and mass distributions. Being aware of this helps us predict how quickly an object can start or stop spinning, which is important in many mechanical systems.
- Higher moment of inertia means more force is required to change the object's rotational speed.
- For a solid cylinder, this resistance is captured by \( I = \frac{1}{2}MR^2 \).
\[ TR = I \alpha \]
The solution of this expression highlights how the cylinder's distribution of mass affects its rotational movement. Moments of inertia are unique for different shapes and mass distributions. Being aware of this helps us predict how quickly an object can start or stop spinning, which is important in many mechanical systems.
power in mechanical systems
Power in mechanical systems is the rate at which work is done or energy is transferred. It can be calculated as the product of force and velocity. In this exercise, the power \( P \) developed by gravitational force is:
\[ P = Fv \]
Where:
\[ P = Mg \left( \frac{2}{3}gt \right) = \frac{2}{3}Mg^2t \]
This shows the continuous energy conversion from potential to kinetic as the cylinder descends. Understanding power, and its applications, enables us to design efficient mechanical systems that effectively utilize energy over time.
\[ P = Fv \]
Where:
- Force \( F \) is the weight \( Mg \).
- Velocity \( v \) is found with \( a \cdot t \), where \( a = \frac{2}{3}g \).
\[ P = Mg \left( \frac{2}{3}gt \right) = \frac{2}{3}Mg^2t \]
This shows the continuous energy conversion from potential to kinetic as the cylinder descends. Understanding power, and its applications, enables us to design efficient mechanical systems that effectively utilize energy over time.
