Rotational Inertia and Tension

[kk727]

Homework Statement

This is a question from the 1999 AP Physics C test.

You can view the picture here, on Page 6:
http://vhphysics.com/ap/x-test-q/Resource/b)%20Physics%20C%20Materials%20(656)/c)%20AP%20C%20Free-Response%20Problems%20(213)/Physics%20C%201999%20Free%20Response.PDF

As shown above, a uniform disk is mounted to an axle and is free to rotate without friction. A thin uniform rod is rigidly attached to the disk, so that it will rotate with the disk. A block is attached to the end of the rod. Properties of the disk, rod, and block are as follows:

Disk: mass=3m, radius=R, moment of inertia about center I_D=1.5mR²
Rod: mass=m, length=2R, moment of inertia about one end I_R=4/3(mR²)
Block: mass=2m

The system is held in equilibrium with the rod at an angle θ to the vertical, as shown above, by a horizontal string of negligible mass with one end attached to the disk and the other to a wall. Express your answers in terms of m, R, θ, and g.

a. Determine the tension in the string.

The string is now cut, and the disk-rod-block system is free to rotate.
b. Determine the following for the instant immediately after the string is cut.
i. The magnitude of the angular acceleration of the disk.
ii. The magnitude of the linear acceleration of the mass at the end of the rod

As the disk rotates, the rod passes the horizontal position shown above.
c. Determine the linear speed of the mass at the end of the rod for the instant the rod is in the horizontal position.

Homework Equations

Not even sure.

The Attempt at a Solution

Again, the pictures are on that website.

I looked at this problem and my brain just wanted to explode. While we learned about inertia, tension, etc. in class, we learned about each individually, and not this in-depth. Trying to put everything together and understand it both conceptually and mathematically has been extremely difficult for me lately.

The solution to this question begins here, on Page 7:
http://vhphysics.com/ap/x-test-q/Resource/b)%20Physics%20C%20Materials%20(656)/d)%20AP%20C%20Rubrics%20(329)/Physics%20C%201999%20Scoring%20Guide.PDF

I looked it over, trying to follow the math, but I still do not understand what the heck is going on here. I get that the forces in the rod and block are going to equal the tension in the string - because it's not moving. But everything else is simply beyond me.

For example, that website says the tension in the rod is mgRsinθ. Why wouldn't it be 2R, since that's what the given length of the rod is?

I'm just extremely confused, and any explanations would be sooooo appreciated! Thanks in advance!

 

[Beaker87]

Part a:

I get that the forces in the rod and block are going to equal the tension in the string

This is not quite correct; the correct statement would be: The TORQUES in the rod and block are equal to the TORQUE applied by the string.

So, we need to find the torque applied by each object in the system.

The disk applies zero torque because it is centered around the axle.

The block applies torque equal to the component of its gravitational force perpendicular to the radius times the distance from the center.

Since the rod has a uniform density, its center of mass is at its center, therefore you can solve this problem by assuming that all of the rod's mass is at its center, R. The rod applies torque equal to the component of its gravitational force perpendicular to the radius times R.

Now you need to add up all of the torques, and the torque applied clockwise by the string will be equal to the sum of the other torques.

Once you have the torque the string applies, you divide it by the radius to get the tension in the string.
*NOTE: If the string were not perpendicular to the radius, you would only wind up with the perpendicular component of the tension.*

Part b:
First, calculate the moment of inertia of the block using the formula: I=Σmr²

Next add up all of the moments of inertia in order to get the total moment of inertia of the system.

Then you can use the formula: τ=Iα with the torque you found in part a in order to find the angular acceleration.

Finally, use the relation: a=rα to find the linear acceleration of the block.
(not on your formula sheet, although it can be derived from v=rω by differentiating both sides with respect to time)

Part c:
The easiest method for this part is to use conservation of energy. Set the ending conditions (θ=π/2) height to zero, and find the height of the rod and the block as a function of θ.
*Note: the height of the rod will be the height of the center of mass of the rod*

Next, use U_g=mgh with your formulas for height to find the initial potential energy of each of these objects (the final potential energy will be zero as you defined the final height to be zero)

Add the two potential energies to get the total potential energy.

The initial potential energy is equal to the final kinetic energy due to the law of conservation of energy, and due to the fact that the final potential energy is zero.

Now solve for ω with the formula: K=(1/2)Iω².

Finally, use v=rω to find the linear velocity.

 

[kk727]

Thank you so much! That's exactly what I needed. Sometimes seeing the solution is enough to help me figure something out, but this wasn't the case. I think I understand what you're saying.

I forgot that you would use the center of mass - for some reason I was looking at the length. That's why the rod was just R instead of 2R?

For Part C, are you just ignoring the disk? Because we're just focusing on the rod and block rotating? Also, we talked about this briefly in class, and my teacher said that the initial height of the rod would be rcosθ, because it starts up at some angle (θ). The height of the block was then 2rcosθ. Is this what you were saying or were we completely wrong?

Again, thank you!

 

[kk727]

Nevermind, I pretty much understand it! Doing similar problems just helped everything connect in my head! Thank you for all of your explanations though, it helped me so much!

 

[rwisz]

First of all, it is completely normal to feel overwhelmed and confused when trying to solve a complex physics problem like this. It takes time and practice to fully understand and apply all the concepts involved. It is important to break down the problem into smaller parts and understand each part before putting it all together.

To start, let's look at part a of the problem. We are given the mass and moment of inertia of the disk, rod, and block. We are also told that the system is in equilibrium, which means the forces acting on it must be balanced. The only two forces acting on the system are the tension in the string and the weight of the block (mg). Since the system is not moving, we know that the net force must be zero. This means that the tension in the string must be equal in magnitude to the weight of the block, but in the opposite direction.

Now, let's move on to part b. We are asked to determine the angular and linear accelerations of the system immediately after the string is cut. When the string is cut, the only force acting on the system is the weight of the block. This force will cause the system to rotate and accelerate. The angular acceleration of the disk can be found using Newton's second law for rotational motion: τ=Iα, where τ is the net torque, I is the moment of inertia, and α is the angular acceleration. In this case, the only torque acting on the system is due to the weight of the block, and it is given by τ=mgRsinθ. This torque causes the disk to rotate with an angular acceleration of α=τ/I=mgRsinθ/I=2mgRsinθ/3mR^2=2gsinθ/3R.

To find the linear acceleration of the block at the end of the rod, we can use Newton's second law for linear motion: F=ma. The only force acting on the block is the weight of the block, and it is given by F=mg. Since the block is attached to the rod, it will also experience a centripetal force directed towards the center of rotation. This centripetal force is provided by the tension in the rod, and it is given by F=mv^2/R, where v is the linear speed of the block. Setting these two forces equal, we get mg=mv^2/R, which can be rearranged