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Problems and Solutions on Rotational Dynamics

# Article: Problems and Solutions on Rotational Dynamics

## Understanding Rotational Dynamics

Rotational dynamics refers to the part of mechanics that deals with the rotation of objects and the forces responsible for it. It is a central concept in classical mechanics, laying the foundation for analyzing the motion of rigid bodies. Several problems in this domain involve understanding torque, rotational inertia, angular momentum, and the role of external and internal forces in rotational motion.

## Common Problems in Rotational Dynamics

Problems in rotational dynamics can be complex due to the involvement of various physical quantities and principles. Among the most common issues are:

1. Calculating the torque required for a specific angular acceleration.
2. Determining the moment of inertia for various geometrical bodies.
3. Analyzing the angular momentum conservation in isolated systems.
4. Solving for forces in static equilibrium situations involving torque.
5. Finding the relationship between linear and angular quantities for rolling motion without slipping.

## Solutions to Rotational Dynamics Problems

Solving problems in rotational dynamics generally requires employing Newton’s second law for rotation and its related concepts. To approach these issues effectively, consider the following strategies:

– **Understand the rotation axis**: Specify the axis of rotation, as this will impact the moment of inertia and torque.
– **Apply conservation laws**: Utilize the conservation of angular momentum when the net external torque acting on a system is zero.
– **Use the right equations**: Employ the rotational kinematics equations and combine them appropriately with linear motion equations where necessary.
– **Analyze forces carefully**: For statics problems, understand the conditions for equilibrium, i.e., net force and net torque must both be zero.

## Example Problems and Solutions in Rotational Dynamics

Here are 20 problems with solutions related to rotational dynamics, involving a combination of theory and calculations.

### Problem 1
A solid disc rotates with an angular velocity of \( \omega = 10 \ \text{rad/s} \). If its moment of inertia is \( I = 2 \ \text{kg.m}^2 \), what is its angular momentum?

**Solution:**
Angular momentum \( L \) can be calculated using the formula \( L = I\omega \).
\[ L = 2 \ \text{kg.m}^2 \cdot 10 \ \text{rad/s} = 20 \ \text{kg.m}^2/\text{s} \].

### Problem 2
What force would be necessary to maintain a torque of \( 60 \ \text{N.m} \) on a circular wheel of radius \( 2 \ \text{m} \)?

**Solution:**
Torque \( \tau \) is given by \( \tau = rF\sin(\theta) \), where \( F \) is the force applied, \( r \) is the radius, and \( \theta \) is the angle between the force vector and the lever arm. Given that the force is applied perpendicularly, \( \sin(\theta) = 1 \), hence
\[
F = \frac{\tau}{r} = \frac{60 \ \text{N.m}}{2 \ \text{m}} = 30 \ \text{N}.
\]

### Problem 3
A figure skater with an initial angular speed of \( 3 \ \text{rad/s} \) pulls in their arms, reducing their moment of inertia from \( 4 \ \text{kg.m}^2 \) to \( 2 \ \text{kg.m}^2 \). What is the skater’s final angular speed?

**Solution:**
Since there are no external torques, angular momentum \( L \) is conserved.
\[
L_{\text{initial}} = L_{\text{final}} \Rightarrow I_{\text{initial}}\omega_{\text{initial}} = I_{\text{final}}\omega_{\text{final}}
\]
\[
4 \ \text{kg.m}^2 \cdot 3 \ \text{rad/s} = 2 \ \text{kg.m}^2 \cdot \omega_{\text{final}}
\]
\[ \omega_{\text{final}} = \frac{4 \ \text{kg.m}^2 \cdot 3 \ \text{rad/s}}{2 \ \text{kg.m}^2} = 6 \ \text{rad/s} \].

### Problem 4
A solid sphere of mass \( 5 \ \text{kg} \) and radius \( 0.5 \ \text{m} \) is rolling without slipping. If its center of mass has a linear velocity of \( 2 \ \text{m/s} \), what is its angular velocity?

**Solution:**
For rolling without slipping, \( v = r\omega \), where \( v \) is the linear velocity and \( \omega \) is the angular velocity.
\[ \omega = \frac{v}{r} = \frac{2 \ \text{m/s}}{0.5 \ \text{m}} = 4 \ \text{rad/s} \].

### Problem 5
What is the moment of inertia \( I \) of a rod with mass \( m = 3 \ \text{kg} \) and length \( L = 2 \ \text{m} \) about an axis perpendicular to the rod at one end?

**Solution:**
The moment of inertia of a rod about an axis perpendicular to its length and through one end is \( I = \frac{1}{3}mL^2 \).
\[ I = \frac{1}{3}(3 \ \text{kg})(2 \ \text{m})^2 = 4 \ \text{kg.m}^2 \].

### Problem 6
A torque of \( 12 \ \text{N.m} \) is applied to a wheel for \( 4 \ \text{s} \). If the wheel starts from rest, what is its final angular velocity, given that its moment of inertia is \( 3 \ \text{kg.m}^2 \)?

**Solution:**
Using Newton’s second law for rotation \( \tau = I\alpha \), where \( \tau \) is the torque and \( \alpha \) is the angular acceleration.
\[
\alpha = \frac{\tau}{I} = \frac{12 \ \text{N.m}}{3 \ \text{kg.m}^2} = 4 \ \text{rad/s}^2.
\]
Since the wheel starts from rest, \( \omega_{\text{final}} = \alpha t \).
\[ \omega_{\text{final}} = 4 \ \text{rad/s}^2 \cdot 4 \ \text{s} = 16 \ \text{rad/s} \].

### Problem 7
A potter’s wheel with a radius of \( 0.3 \ \text{m} \) is spinning at \( 50 \ \text{rad/s} \). The potter presses down on the wheel and brings the system to a halt in \( 5 \ \text{s} \). What average torque did the potter apply if the moment of inertia is \( 0.6 \ \text{kg.m}^2 \)?

**Solution:**
We first need to find the angular deceleration \( \alpha \) using \( \omega_f = \omega_i + \alpha t \).
\[
0 \ \text{rad/s} = 50 \ \text{rad/s} + \alpha (5 \ \text{s})
\]
\[
\alpha = -\frac{50 \ \text{rad/s}}{5 \ \text{s}} = -10 \ \text{rad/s}^2.
\]
Now, we calculate the torque using \( \tau = I\alpha \).
\[
\tau = (0.6 \ \text{kg.m}^2)(-10 \ \text{rad/s}^2) = -6 \ \text{N.m}
\]
(Note the negative sign indicates that the torque acts to reduce the angular velocity.)

### Problem 8
A cylindrical pulley with a mass of \( 2 \ \text{kg} \) and a radius of \( 0.5 \ \text{m} \) is used to lift a \( 10 \ \text{kg} \) object. What is the minimum force required to lift the object if the system starts from rest?

**Solution:**
First, we find the moment of inertia of the pulley (a cylinder) using \( I = \frac{1}{2}mR^2 \).
\[ I = \frac{1}{2}(2 \ \text{kg})(0.5 \ \text{m})^2 = 0.25 \ \text{kg.m}^2 \]
Since the system starts from rest, we need a force \( F \) that at least matches the gravitational force acting on the object to start lifting it.
\[ F_{\text{gravity}} = m_{\text{object}}g = (10 \ \text{kg})(9.81 \ \text{m/s}^2) = 98.1 \ \text{N} \]
The minimum force \( F \) required must then satisfy \( F = F_{\text{gravity}} \).
\[ F_{\text{min}} = 98.1 \ \text{N} \]

### Problem 9
A door is pushed open with a force of \( 10 \ \text{N} \) applied at a distance of \( 0.8 \ \text{m} \) from the hinges. What is the resulting torque about the hinges?

**Solution:**
Torque \( \tau \) is calculated by \( \tau = rF\sin(\theta) \). If the force is applied perpendicular to the door, \( \theta = 90^\circ \) and \( \sin(90^\circ) = 1 \).
\[ \tau = (0.8 \ \text{m})(10 \ \text{N}) = 8 \ \text{N.m} \]

### Problem 10
A gyroscope wheel has an angular momentum of \( 0.4 \ \text{kg.m}^2/\text{s} \) and a moment of inertia of \( 0.1 \ \text{kg.m}^2 \). What is the angular velocity of the gyroscope wheel?

**Solution:**
Angular momentum is related to the moment of inertia and angular velocity by \( L = I\omega \).
\[ \omega = \frac{L}{I} = \frac{0.4 \ \text{kg.m}^2/\text{s}}{0.1 \ \text{kg.m}^2} = 4 \ \text{rad/s} \]

### Problem 11
Calculate the kinetic energy of a rotating disk with an angular velocity of \( 15 \ \text{rad/s} \) and a moment of inertia of \( 1.2 \ \text{kg.m}^2 \).

**Solution:**
The rotational kinetic energy (\( KE \)) of an object is given by \( KE = \frac{1}{2}I\omega^2 \).
\[ KE = \frac{1}{2}(1.2 \ \text{kg.m}^2)(15 \ \text{rad/s})^2 = 135 \ \text{J} \]

### Problem 12
A child sits at a distance of \( 2 \ \text{m} \) from the center of a merry-go-round. If the child pulls herself towards the center to a distance of \( 1 \ \text{m} \), and her mass is \( 20 \ \text{kg} \), how does this affect the angular velocity if it was initially \( 2 \ \text{rad/s} \)?

**Solution:**
Let’s assume the merry-go-round is initially rotating without the child so that the system’s moment of inertia doesn’t change significantly when the child moves towards the center.

Angular momentum is conserved, so initially \( L_i = I_i \omega_i \), and finally \( L_f = I_f \omega_f \).

The child’s initial moment of inertia is \( I_i = mr^2 = (20 \ \text{kg}) (2 \ \text{m})^2 = 80 \ \text{kg.m}^2 \), and finally \( I_f = (20 \ \text{kg}) (1 \ \text{m})^2 = 20 \ \text{kg.m}^2 \). So,
\[ L_i = L_f \Rightarrow I_i \omega_i = I_f \omega_f \]
\[ (80 \ \text{kg.m}^2)(2 \ \text{rad/s}) = (20 \ \text{kg.m}^2) \omega_f \]
\[ \omega_f = \frac{(80 \ \text{kg.m}^2)(2 \ \text{rad/s})}{20 \ \text{kg.m}^2} = 8 \ \text{rad/s} \]

### Problem 13
A torque of \( 24 \ \text{N.m} \) accelerates a wheel from rest to \( 120 \ \text{rad/s} \) in \( 10 \ \text{s} \). What is the moment of inertia of the wheel?

**Solution:**
First, find the angular acceleration using \( \omega_f = \omega_i + \alpha t \).
\[ 120 \ \text{rad/s} = 0 + \alpha (10 \ \text{s}) \]
\[ \alpha = \frac{120 \ \text{rad/s}}{10 \ \text{s}} = 12 \ \text{rad/s}^2 \]
Now, find the moment of inertia with \( \tau = I\alpha \).
\[ I = \frac{\tau}{\alpha} = \frac{24 \ \text{N.m}}{12 \ \text{rad/s}^2} = 2 \ \text{kg.m}^2 \]

### Problem 14
A turntable with a moment of inertia of \( 0.5 \ \text{kg.m}^2 \) spins at \( 33 \ \text{rpm} \). How much torque is required to stop it in \( 6 \ \text{s} \)?

**Solution:**
First, convert \( 33 \ \text{rpm} \) to \( \text{rad/s} \): \( 33 \ \text{rpm} \) is \( \frac{33}{60} \times 2\pi \ \text{rad/s} \approx 3.455 \ \text{rad/s} \). Calculate the angular deceleration:
\[ \alpha = \frac{\omega_f – \omega_i}{t} = \frac{0 – 3.455 \ \text{rad/s}}{6 \ \text{s}} \approx -0.576 \ \text{rad/s}^2 \]
Now, calculate the torque using \( \tau = I\alpha \):
\[ \tau = (0.5 \ \text{kg.m}^2)(-0.576 \ \text{rad/s}^2) \approx -0.288 \ \text{N.m} \]

### Problem 15
Two children, each with a mass of \( 30 \ \text{kg} \), sit at opposite ends of a see-saw. They are \( 3 \ \text{m} \) from the pivot point. What is the net torque about the pivot point?

**Solution:**
Since the children are at equal distances and have equal masses, their torques about the pivot will cancel each other out.
\[ \tau_1 = r_1F_1 = (3 \ \text{m})(30 \ \text{kg} \cdot 9.8 \ \text{m/s}^2) = 882 \ \text{N.m} \]
\[ \tau_2 = r_2F_2 = (3 \ \text{m})(30 \ \text{kg} \cdot 9.8 \ \text{m/s}^2) = 882 \ \text{N.m} \]
The net torque is \( \tau_{\text{net}} = \tau_1 – \tau_2 = 0 \ \text{N.m} \)

### Problem 16
A hollow cylinder (moment of inertia \( I = mr^2 \)) of radius \( 0.4 \ \text{m} \) and mass \( 3 \ \text{kg} \) rolls down an incline without slipping. If the cylinder starts from rest and covers a distance of \( 5 \ \text{m} \) down the slope in \( 2 \ \text{s} \), what is the final velocity of its center of mass?

**Solution:**
For rolling without slipping, \( a = r\alpha \) where \( a \) is the linear acceleration and \( \alpha \) is the angular acceleration. Since \( v = at \), we need to find \( a \) first.
\[ s = \frac{1}{2}at^2 \Rightarrow 5 \ \text{m} = \frac{1}{2}a(2 \ \text{s})^2 \Rightarrow a = \frac{5 \ \text{m}}{2 \ \text{s}^2} = \frac{5 \ \text{m}}{4 \ \text{s}^2} \]
So the final velocity of the center of mass \( v = at \):
\[ v = \left(\frac{5 \ \text{m}}{4 \ \text{s}^2}\right)(2 \ \text{s}) = \frac{5}{2} \ \text{m/s} \]

### Problem 17
What is the torque required to provide a \( 4 \ \text{kg} \) solid sphere with radius \( 0.3 \ \text{m} \) an angular acceleration of \( 2 \ \text{rad/s}^2 \)?

**Solution:**
The moment of inertia for a solid sphere is \( I = \frac{2}{5}mr^2 \).
\[ I = \frac{2}{5}(4 \ \text{kg})(0.3 \ \text{m})^2 = 0.144 \ \text{kg.m}^2 \]
Now calculate the torque:
\[ \tau = I\alpha = (0.144 \ \text{kg.m}^2)(2 \ \text{rad/s}^2) = 0.288 \ \text{N.m} \]

### Problem 18
A car wheel of radius \( 0.2 \ \text{m} \) accelerates from \( 0 \ \text{rad/s} \) to \( 100 \ \text{rad/s} \) in \( 5 \ \text{s} \). If the wheel’s moment of inertia is \( 0.8 \ \text{kg.m}^2 \), calculate the torque that caused this acceleration.

**Solution:**
Find the angular acceleration \( \alpha \):

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