Angular Momentum & Newton's Second Law - Online Tutor, Practice Problems & Exam Prep

Angular momentum (L) relates to the rotational version of Newton's second law, where the sum of torques equals the change in angular momentum over time. The equation can be expressed as $\sum \tau =\frac{∆}{L}/∆t$. In a practical example, applying a torque of 80 N·m to a 10 kg object spinning at 180 RPM with a radius of 5 m results in a stopping time of approximately 59 seconds, illustrating the conservation of angular momentum and the effects of torque.

Angular Momentum & Newton's Second Law

Hey, guys. So in this video, I want to show you how there's a relationship between angular momentum, l, and the rotational version of Newton's second law. Let's check it out. So remember that Newton's second law, the linear version, the first one you learn, f equals ma, can be rewritten in terms of linear momentum. Instead of f equals ma, I can write the sum of all forces equals delta p, change in momentum over delta t, change in time or time elapsed. The rotational version of Newton's second law, sum of all torques equals I alpha can also be rewritten, but in terms of angular instead of linear momentum because it's the rotational version. So instead of sum of forces, I'm going to have sum of torques. And instead of delta p, I'm going to have delta l. L is the equivalent, the rotational version, if you will, of p, l. And this is just time. Alright. Let's do a quick example to see how we would use this equation to solve a problem.

So I have a small object, small means it's going to be a point mass. Okay. 10 kilograms in mass, mass equals 10, spins with an RPM of 180 so RPM equals 180 in a circular path of radius 5. Now the radius of a circular path is really the distance to the center, so that's little r equals 5 and you have an m here and the object is going this way with some Omega. It says that you apply a constant torque of 80, so torque equals 80. In trying to stop it, so you are applying if the object, is spinning this way, the question doesn't tell you which way it spins. But let's say it's spinning this way. So you're applying a torque in this direction to try to stop it. I want to know how long will it take for it to come to a complete stop. So I'm going from some Omega initial that is not 0 to an Omega final that is 0. So we're stopping and I want to know how long does it take to do that. Now notice that I'm giving you torque and I'm asking for delta t. Therefore, this equation right here can be used, right, because it's got both of those. That's what we're going to do. Sum of torques equals delta l over delta t. I'm looking for delta t. So I'm going to, I'm going to solve for delta t. I'm going to move it up and I'm going to move so delta l is here and I'm going to move torque down. There's only one torque here which is the 80. So, I'm going to put that in here. Delta l is l final minus l initial and then this torque is 80. L, remember, is I omega. So you can think of this as I, final omega final, I initial omega initial divided by 80. Now we are coming to a stop, so Omega final will be 0. So this whole thing is 0. So we're left with, we're left with, I initial, omega initial, minus, divided by 80. Okay. Technically, I guess I should have put this torque here is causing you to slow down. So it's a negative torque. Okay? It's a negative torque because it's causing you to slow down. Also, the way that I drew it, I have this as a positive because it's clockwise or counterclockwise and torque is this way. So this is going to be negative. It's clockwise and also it's trying to stop you. So I'm going to put a negative here. And I needed that negative to cancel out, otherwise, end up with a negative time. Okay. But that's just a sign thing, not too big of a deal. So what we're going to do now is plug in these numbers. I don't have I, but I can find it because I is going to be the I of a point mass, which is m r squared. M is 10. R is 5. So this is going to be 250. And so I got 250 there. The negatives cancel. And then omega, I have to get omega as well. Omega is 2πf which is 2πrpm over 60. So it's 2 π, 180 divided by 60 is 3. 3 times 2π, 6π. Okay. So 6π goes right here, and we are done. You multiply this whole thing multiply this whole thing, and we get 59 seconds. Okay. 59 seconds. So that's how long it would take, to cause this object to slow down if you apply this torque. Alright? Very straightforward. Just had to plug it into the equation. The only part that you had to sweat a little bit is that, when you expanded I, you had to go find I and then you had to go find omega. But other than that, it was just straightforward plug into the equation. Alright. That's it for this one. Let me know if you have any questions and let's keep going.

A solid disc of mass M = 40 kg and radius R = 2 m is free to rotate about a fixed, frictionless, perpendicular axis through its center. You apply a constant, tangential force on the disc's surface (as shown), to get it to spin. Calculate the magnitude of the force needed to get the disc to 100 rad/s in just one minute.

• A. 33.3 N
• B. 66.7 N
• C. 133 N
• D. 167 N

Here’s what students ask on this topic:

What is the relationship between angular momentum and Newton's second law?

The relationship between angular momentum (L) and Newton's second law in its rotational form is given by the equation:

$\sum \tau =\frac{∆L}{∆t}$

Here, the sum of torques (Στ) is equal to the change in angular momentum (ΔL) over the change in time (Δt). This is analogous to the linear version of Newton's second law, where the sum of forces equals the change in linear momentum over time. In essence, torque is to angular momentum what force is to linear momentum.

How do you calculate the stopping time of a rotating object using torque?

To calculate the stopping time of a rotating object using torque, you can use the equation:

$\sum \tau =\frac{∆L}{∆t}$

Rearrange to solve for Δt:

$∆t=\frac{∆L}{\sum \tau }$

First, find the initial angular momentum (L_initial) using L = Iω, where I is the moment of inertia and ω is the angular velocity. Then, calculate the change in angular momentum (ΔL = L_final - L_initial). Finally, divide ΔL by the applied torque (Στ) to find the stopping time.

What is the moment of inertia for a point mass?

The moment of inertia (I) for a point mass is calculated using the formula:

$I=m{r}^2$

Here, m is the mass of the object and r is the distance from the axis of rotation. This formula is crucial for solving problems involving rotational motion, as it allows you to determine the resistance of the object to changes in its rotational state.

How do you convert RPM to angular velocity?

To convert RPM (revolutions per minute) to angular velocity (ω) in radians per second, use the formula:

$\omega =\frac{2\pi \mathrm{RPM}}{60}$

Here, RPM is the rotational speed in revolutions per minute. Multiply RPM by 2π to convert to radians per minute, then divide by 60 to convert to radians per second. This conversion is essential for solving problems involving angular motion.

What is the significance of torque in rotational motion?

Torque (τ) is a measure of the rotational force applied to an object. It is significant in rotational motion because it determines the rate of change of angular momentum. The equation:

$\sum \tau =\frac{∆L}{∆t}$

shows that the sum of torques is equal to the change in angular momentum over time. This relationship is analogous to how force affects linear momentum in linear motion. Torque is crucial for understanding how rotational systems respond to applied forces.