Angular Velocity and Acceleration - AP Physics 1

The angular velocity should not change based on the radius of the second hand. No matter what size the second hand, it will still cover one revolution every minute or 60s. The linear velocity will be greater and the angular momentum will also be greater for the clocktower, but its angular velocity will be the same. This can be seen by looking at the equation for angular velocity:

The definition of angular velocity is .

By identifying the given information to be and , we can plug this into the equation to calculate the angular velocity:

You may recall the kinematic equation that relates final velocity, initial velocity, acceleration, and distance, respectively:

Well, for rotational motion (such as in this problem), there is a similar equation, except it relates final angular velocity, intial angular velocity, angular acceleration, and angular distance, respectively:

The wheel starts at rest, so the initial angular velocity, , is zero. The total number of revolutions of the wheel is given to be 5 revolutions. Each revolution is equivalent to an angular distance of radians. So, we can convert the total revolutions to an angular distance to get:

The final angular velocity was given as in the text of the question. So, we should use the above equation to solve for the angular acceleration, .

For a rotating object, or an object moving in a circular path, the relationship between angular acceleration and linear acceleration is

Linear acceleration is given by , angular acceleration is , and the radius of the circular path is .

For circular/centripetal motion, the linear acceleration is related to the object's linear velocity by

We know the linear velocity is , and the radius is 1.5 m, so we can find the linear acceleration...

Now that we have the linear acceleration, we can use this in the equation at the top to find the angular acceleration...

The angular velocity of the second hand of a clock can be found by dividing the number of radians the second hand will travel over a known period of time. Thankfully for a clock, we know that the second hand will make one revolution, i.e. covering in one minute, or 60s. The formula for angular velocity is:

So the angular velocity, is , which simplifies to our answer,

Angular velocity, in , is given by the length traveled divided by the time taken to travel the length:

We are told that the amount of time taken to make one revolution is 3min. One revolution is equal to , and 3 minutes is equal to 180 seconds. Divide the radian value by the seconds value to get the angular velocity.

For this question, the angular velocity can be given by the equation:

, where is the angle made and is the time taken to make this angle.

In this problem, the wheel makes one full revolution() in seconds.

Therefore:

Given initial angular velocity, angular acceleration, and time we can easily solve for final angular velocity with:

This is a centripetal force problem. It's important to know that when you draw your free body diagram (FBD), centripetal force () isn't drawn, similar to how net forces () aren't drawn. is actually the centripetal or circular form of. So when we talk about circular motion such as this, we can set the two equal to each other. Why is this? Well we are trying to translate this object's circular motion into it's linear velocity (our answer).

We can now set, or equal to the combination of vertical forces in our FBD

But like we said earlier, we need to connect our linear forces to our centripetal forces. Remember this equation?

Now let's put it all together

Time to pull in some algebra now. Remember we want the speed of the bucket, or the velocity of the bucket at the lowest point of the circle, so solve for :

If you plut in everything and solve (remember to use PEMDAS), your answer should be

Centripetal acceleration is as follows:

All we need to do is solve and plug in for

= distance traveled and t is the time it took to travel that distance

We know that the yo-yo made 4 revolutions in .

One revolution is the circumference of the yo-yo path:

So we can either solve for in terms of how far the yo-yo went in one second (I), or how long it took to make one revolution (II); I'll show both, either one is correct if you wish to plug in

or

REMEMBER: Both of these will give you the same answer (check it out if you're good at algebra). Physics is all about comfort in manipulation so choose the way that suits you!

Now plug this into our equation. Lucky for you, the radius was

Note: I put in the units so you can see how they cancel out algebraically

If the ferris wheel has height then it must have radius .

The circumference of the ferris wheel, or the distance of one rotation, is then:

Convert the given velocity into meters per minute, or :

Find rotations per minute:

Convert to :

Convert to

Use the following relationship and plug in known values:

Convert units of time into radians per second:

Convert to linear distance:

Combine equations:

Convert to meters and seconds and plug in values:

We can use the expression for conservation of energy to solve this problem:

Our initial state will be when the rod is horizontal, and our final state will be when mass A is at its highest point. If we assume that point p is at a height of 0 and the system is at rest and mass A is at its highest point, we can eliminate initial potential energy and final kinetic energy to get:

Expanding these terms and applying them to both masses, we get:

We don't need to separate the velocity components for each mass since they are always traveling at the same speed. Since the masses are attached to a rigid rod that spins around its midpoint, we know that the heights of the two masses (with respect to point p) will be equal and opposite. In expression form:

Substituting this into our equation, we get:

Rearranging for final height, we get:

We have values for all of these variables, so time to plug and chug:

Now we can use the sine function to determine what the angle c is at this point:

Where the opposite side is the height we just calculated and the hypotenuse is half the length of the rod. Therefore, we get:

Taking the inverse sine of both sides, we get:

Substituting in our values, we get:

Let's begin with the expression for conservation of energy:

The problem statement tells us that the sphere is initially at rest, so we can eliminate initial kinetic energy. Also, if we assume that the height at the bottom of the ramp is 0, we can eliminate final potential energy. We then have:

Then we can expand both of these terms. We need to remember that kinetic energy will have both a linear and rotational aspect. We then get equation (1):

Moving from left to right, let's begin substituting in expressions for unknown variables. The first term we don't know is height. However, we can use the length of the slope and the its angle to determine the height at the top of the ramp:

Rearranging for height, we get equation (2):

The next term we don't know is final velocity. We can use the relationship between angular and linear velocity:

Rearranging for linear velocity, we get equation (3):

Moving on, the next term we don't know is moment of inertia. We will use the expression for a sphere to get equation (4):

The last term is final rotational velocity. However, this is what we're solving for, so we'll leave it alone. Now let's substitute equations 2, 3, and 4 back into equation 1:

Multiplying each side the equation by , we get:

Factoring the right side of the equation:

Rearranging for final rotational velocity:

We know each of these values, so time to plug and chug:

The angular velocity is given by:

One circle is equal to , thus

If the cars are racing side by side on a circular track, then they have the same angular velocity, because they complete their circles in the same amount of time.