Conquer Physics: 2 Problems To Ace Your 2nd-Year Studies!
Hey there, physics enthusiasts! Are you ready to flex those brain muscles and tackle some exciting challenges? This article is your ultimate guide to conquering two awesome physics problems perfect for your second-year studies. We'll break down the concepts, provide step-by-step solutions, and ensure you're well-equipped to ace your exams. So grab your pens, paper, and let's dive into the fascinating world of physics! Ready to level up your understanding and boost your problem-solving skills? Let's get started!
Problem 1: Unveiling the Secrets of Rotational Motion
Alright, guys, let's kick things off with a problem that dives headfirst into the intriguing realm of rotational motion. This is where things start to spin – literally! We will explore concepts like angular velocity, torque, and moment of inertia. Here's the scenario:
A uniform solid sphere with a mass of 5 kg and a radius of 0.2 meters is initially at rest. A constant torque of 2 Nm is applied to the sphere, causing it to rotate. Assuming no friction, determine the following:
- (a) The angular acceleration of the sphere.
- (b) The angular velocity of the sphere after 4 seconds.
- (c) The rotational kinetic energy of the sphere after 4 seconds.
Sounds like fun, right? Don't worry, we'll break this down step by step to make it super easy to understand. This problem highlights how torque affects rotational acceleration and how energy is stored in a rotating object. Let's start with part (a).
(a) Cracking the Code of Angular Acceleration
To find the angular acceleration (α), we'll use the fundamental relationship between torque (τ), moment of inertia (I), and angular acceleration:
τ = Iα
First, we need to calculate the moment of inertia (I) of the solid sphere. For a uniform solid sphere, the moment of inertia is given by:
I = (2/5)MR²
Where:
- M = mass of the sphere = 5 kg
- R = radius of the sphere = 0.2 m
Plugging in the values, we get:
I = (2/5) * 5 kg * (0.2 m)² = 0.04 kg·m²
Now, we can rearrange the torque equation to solve for α:
α = τ / I
We know that the applied torque (τ) is 2 Nm. Therefore:
α = 2 Nm / 0.04 kg·m² = 50 rad/s²
So, the angular acceleration of the sphere is 50 rad/s². See? Not so bad, eh?
(b) Speeding Up: Finding the Angular Velocity
Next up, we want to find the angular velocity (ω) after 4 seconds. Since the angular acceleration is constant, we can use a simple kinematic equation:
ω = ω₀ + αt
Where:
- ω₀ = initial angular velocity = 0 rad/s (since the sphere starts from rest)
- α = angular acceleration = 50 rad/s²
- t = time = 4 s
Plugging in the values:
ω = 0 rad/s + (50 rad/s²) * 4 s = 200 rad/s
Therefore, the angular velocity of the sphere after 4 seconds is 200 rad/s. The sphere is really spinning now!
(c) Unleashing Rotational Kinetic Energy
Finally, let's calculate the rotational kinetic energy (KEᵣ) after 4 seconds. The formula for rotational kinetic energy is:
KEᵣ = (1/2)Iω²
Where:
- I = moment of inertia = 0.04 kg·m²
- ω = angular velocity = 200 rad/s
Plugging in the values:
KEᵣ = (1/2) * 0.04 kg·m² * (200 rad/s)² = 800 J
So, the rotational kinetic energy of the sphere after 4 seconds is 800 J. The sphere has stored a significant amount of energy due to its rotation. That's a wrap for Problem 1! We've successfully navigated the twists and turns of rotational motion. You should feel pretty good about your understanding now.
Problem 2: Riding the Waves of Electromagnetism
Now, let's shift gears and dive into the fascinating world of electromagnetism. In this problem, we'll explore the relationship between electric and magnetic fields and the forces they exert. Here's the challenge:
A proton (charge +1.6 x 10⁻¹⁹ C, mass 1.67 x 10⁻²⁷ kg) enters a uniform magnetic field with a velocity of 2 x 10⁶ m/s perpendicular to the field. The magnetic field strength is 0.5 T. Determine the following:
- (a) The magnitude of the magnetic force on the proton.
- (b) The radius of the proton's circular path in the magnetic field.
- (c) The period of the proton's circular motion.
This problem will test your knowledge of the Lorentz force, circular motion, and how magnetic fields affect charged particles. Let's get started. Are you ready to explore magnetic forces? Let's break it down.
(a) Unveiling the Magnetic Force
The magnetic force (F) on a charged particle moving in a magnetic field is given by the following formula:
F = qvBsinθ
Where:
- q = charge of the proton = +1.6 x 10⁻¹⁹ C
- v = velocity of the proton = 2 x 10⁶ m/s
- B = magnetic field strength = 0.5 T
- θ = angle between the velocity and the magnetic field = 90° (since they are perpendicular), and sin(90°) = 1.
Plugging in the values:
F = (1.6 x 10⁻¹⁹ C) * (2 x 10⁶ m/s) * (0.5 T) * 1 = 1.6 x 10⁻¹³ N
Therefore, the magnitude of the magnetic force on the proton is 1.6 x 10⁻¹³ N. This force is what causes the proton to move in a circular path.
(b) Charting the Circular Path: Finding the Radius
The magnetic force acts as the centripetal force, causing the proton to move in a circle. The centripetal force is given by:
F = mv²/r
Where:
- m = mass of the proton = 1.67 x 10⁻²⁷ kg
- v = velocity of the proton = 2 x 10⁶ m/s
- r = radius of the circular path
Since the magnetic force is equal to the centripetal force, we can equate the two formulas:
q * v * B = mv²/r
Rearranging to solve for the radius (r):
r = (mv) / (qB)
Plugging in the values:
r = (1.67 x 10⁻²⁷ kg * 2 x 10⁶ m/s) / (1.6 x 10⁻¹⁹ C * 0.5 T) = 0.04175 m
So, the radius of the proton's circular path is approximately 0.04175 meters. The proton is tracing a beautiful, tiny circle!
(c) Defining the Cycle: Calculating the Period
The period (T) of the circular motion is the time it takes for the proton to complete one full revolution. It's related to the circumference of the circle and the velocity of the proton:
T = 2πr / v
We know the radius (r = 0.04175 m) and the velocity (v = 2 x 10⁶ m/s). Plugging in the values:
T = (2π * 0.04175 m) / (2 x 10⁶ m/s) = 1.311 x 10⁻⁷ s
Therefore, the period of the proton's circular motion is approximately 1.311 x 10⁻⁷ seconds. The proton is zooming around really, really fast!
Final Thoughts: Keep Exploring!
Congratulations, physics enthusiasts! You've successfully solved two challenging problems and expanded your understanding of rotational motion and electromagnetism. Remember, practice is key to mastering these concepts. Keep exploring, keep questioning, and keep having fun with physics. You've got this! Don't be afraid to revisit the concepts, work through more problems, and ask questions. Physics can be challenging, but it is also incredibly rewarding. Keep up the amazing work, and keep exploring the wonders of the physical world! Go out there and impress your professor and classmates. I hope this helps you with your physics studies. Best of luck on your academic journey, and keep up the great work! That's all for today, guys!