Homework Help: Solving Rotation Problems | Physics Forum Tutors

[ubiquinone]

Hi, I read a whole chapter in the textbook on rotation and angular motion. Then I tried to apply those concepts by solving problems, but I can't seem to solve the part B questions. The part A ones which were primarily subbing into the formula, I could solve, but these I am having a lot of trouble. I was wondering if anyone may please demonstrate to me how to solve this problem, so I can solve other ones like this in my practice problems. Thank you so much physics forum tutors!

Question: A yo-yo of a uniform cylinder of radius $$r=8.0mm$$ with $$2$$ uniform cylinders of radius $$R = 30.0 mm$$ placed on each slide of it on a common axis. The small cylinder has mass $$50.0 g$$; each of the large cylinders has a mass of $$250.0 g$$. A string is wrapped around the center small cylinder and pulls the yoyo on a horizontal surface as shown so that it rolls without slipping. The tension $$F_T = 3.0 N$$ a) the rotational inertia of the yo-yo b) the linear acceleration of the center of mass c) The $$F_f$$ acting on the yoyo d) the minimum coefficient of static friction if the yo-yo rolls but doesn't slide on the surface.

I think I can solve a) and b), here goes
The two disks share a common axis so their center of mass coincides. Can we say that $$(m_1+m_2)a=F_T$$ or $$a=\frac{3.0N}{(.25+.05)kg}=10m/s^2$$
So angular acceleration is $$\alpha=\frac{a}{r}={10m/s^2}{0.008m}=1250rad/s$$
Then, $$rF=I\alpha$$ so $$I=\frac{rF}{\alpha}$$?

 

[OlderDan]

You need a picture, or at least a description of the location and direction of the string in the problem. You are missing the mass of one of the big disks, and you have apparently left out the contribution from friction.

 

[andrevdh]

I would think that one need to consider the tension in the string as applying a torque to the yy. That is the string is rotating the yy about its point of contact with the surface. You also need to realize that the line of action of the tension do not go through the com of the yy.

Once you've got the angular acceleration (aa) $$\alpha$$ of the yy you can calculate the acceleration of its com. That is the whole yy experiences the same aa since it is a rigid body. Which enables you the calculate the tangential acceleration of its com using the aa, which will be the linear acceleration of its com.

 

[ubiquinone]

Hi thanks guys for replying to my question! I'm going to give it another try.

How bout first solving for the moment of inertial of the yo-yo:
Let the small cylinder be $$m_1=0.05kg$$ and each of the big cylinders be $$m_2=0.25kg$$. Also, let the radius of the small cylinder be $$r=0.008m$$ and the radius of the big cylinders be $$R=0.03m$$
Therefore, a) $$I=m_1r^2+2MR^2=(0.05kg)(0.008m)^2+2(0.25kg)(0.03m)^2=4.532\times 10^{-4}kg\cdot m^2$$
b) $$\tau=I\alpha$$ or $$\alpha=\frac{rF_T}{I}=\frac{(0.008m)(3.0N)}{4.532\times 10^{-4}kg\cdot m^2}=52.96 rad/s^2$$
So the linear acceleration, $$a=R\alpha$$ (Do I use the radius of the big cylinder?)
If so, $$a=(0.03m)(52.96 rad/s^2)=1.59m/s^2$$
c) Can I use Newton's second law for finding the frictional force?
If so, $$F_{net}=(2M+m_1)a=F_T-F_f$$
Therefore, $$F_f=3.0N - (0.05kg + 2\times 0.25kg)(1.59m/s^2)=2.13N$$.

Are these correct? I'm hoping someone can please check over my work and give me some hints on how to do part d) as well. Thanks again!

 

[andrevdh]

Place a ruler flat on the table. Roll a coin with ridges along the ruler. Yes the coin is rotating about its com, but while it is doing that it is moving forward. So what is actually happening is that the coin is rolling over the point of contact with the ruler. So the tension is causing the yy to rotate about the point of contact with the surface. You therefore need to calculate the I's about this point (use the parallel axis theorem). According to my knowledge

$$I_{disc} = \frac{1}{2}MR^2$$

for an axis perpendicular to the flat sides running through its com.

When you calculate the torque you need to take the distance from just above the small cylinder up to the contact point.

 

[OlderDan]

Hi thanks guys for replying to my question! I'm going to give it another try.

How bout first solving for the moment of inertial of the yo-yo:
Let the small cylinder be $$m_1=0.05kg$$ and each of the big cylinders be $$m_2=0.25kg$$. Also, let the radius of the small cylinder be $$r=0.008m$$ and the radius of the big cylinders be $$R=0.03m$$
Therefore, a) $$I=m_1r^2+2MR^2=(0.05kg)(0.008m)^2+2(0.25kg)(0.03m)^2=4.532\times 10^{-4}kg\cdot m^2$$
b) $$\tau=I\alpha$$ or $$\alpha=\frac{rF_T}{I}=\frac{(0.008m)(3.0N)}{4.532\times 10^{-4}kg\cdot m^2}=52.96 rad/s^2$$
So the linear acceleration, $$a=R\alpha$$ (Do I use the radius of the big cylinder?)
If so, $$a=(0.03m)(52.96 rad/s^2)=1.59m/s^2$$
c) Can I use Newton's second law for finding the frictional force?
If so, $$F_{net}=(2M+m_1)a=F_T-F_f$$
Therefore, $$F_f=3.0N - (0.05kg + 2\times 0.25kg)(1.59m/s^2)=2.13N$$.

Are these correct? I'm hoping someone can please check over my work and give me some hints on how to do part d) as well. Thanks again!

The moment of inertia calculation is not correct. I still need to know whether the string is above or below the center of teh axle, and what direction the string is being pulled. Is it horizontal, vertical, or somewhere in between?

 

[ubiquinone]

Sorry about that OlderDan. The string is under the center of the axel of the smaller radius. The string is being pulled horizontally to the right.

 

[OlderDan]

Sorry about that OlderDan. The string is under the center of the axel of the smaller radius. The string is being pulled horizontally to the right.

OK Got it. The first thing you need to do is find the correct expression for the moment of inertia. andrevdh posted the correct equation for the moment of inertia of a disk about its axis through the CM perpendicular to the flat side. Since you have three concentric disks (two identical), you can use that to find the moment of inertial of the yo-yo about its axis. He also mentioned finding the moment of inertia about the contact point. You can use that also and for the sequence of questions in the problem, it is good to approach it that way. It can be done using just the moment about the CM, but then you will be finding some answers simultaneously. To find the moment of inertia about the contact point you will need to use the parallel axis theorem.

 

[ubiquinone]

Hi thanks OlderDan for replying. I made a rough diagram too. I hope this helps.

Diagram
         *   *
       *       *
      *         *
           O_________ F_T
      *         *
       *       *
F_f______*____*___________
         SURFACE
The outer circle formed by asterisks represents the larger cylinders
The "O" in the center represents the smaller cylinder
The line under the "O" represents the rope and the tension force.
The line below the outer circle represents the surface and the frictional force acting against the yo-yo.

So a) $$\displaystyle I_y=\frac{1}{2}(0.05kg)(0.008m)^2+(2)\frac{1}{2}(0.25kg)(0.03m)^2=2.266\times 10^{-4}kg\cdot m^2$$

b) I'm still thinking about parts b) and c). Can I do something like this:
torque $$\displaystyle =rF_T-RF_f=I_y\alpha$$ (1)
while (net force)=$$F_T-F_f=ma$$ (2)

If I can proceed solving like this, would $$m$$ of equation (2) be the sum of the masses of all three cylinders, (i.e. $$m=(0.25+0.25+0.05)kg=0.55kg?$$)
Also can I rewrite $$a$$ of equation (2) in terms of $$\alpha$$ if so, would $$a=r\alpha$$ or $$a=R\alpha$$.

I would appreciate greatly if anyone can give me more assistance. Again, thank you!

 

[OlderDan]

Hi thanks OlderDan for replying. I made a rough diagram too. I hope this helps.

So a) $$\displaystyle I_y=\frac{1}{2}(0.05kg)(0.008m)^2+(2)\frac{1}{2}(0.25kg)(0.03m)^2=2.266\times 10^{-4}kg\cdot m^2$$

b) I'm still thinking about parts b) and c). Can I do something like this:
torque $$\displaystyle =rF_T-RF_f=I_y\alpha$$ (1)
while (net force)=$$F_T-F_f=ma$$ (2)

If I can proceed solving like this, would $$m$$ of equation (2) be the sum of the masses of all three cylinders, (i.e. $$m=(0.25+0.25+0.05)kg=0.55kg?$$)
Also can I rewrite $$a$$ of equation (2) in terms of $$\alpha$$ if so, would $$a=r\alpha$$ or $$a=R\alpha$$.

I would appreciate greatly if anyone can give me more assistance. Again, thank you!

I did not run my calculator, but the equation is correct for I and that is the I about the CM. Your equations for b) look good, and it is the total mass. Your α and a relationship is valid for R. The reason it is R is because R is the distance from the CM to a point on the horizontal surface where the rim is momentarily pivoting. There is no such point for the inner radius. Also, your equations have been written assuming that α and a are both positive. It is somewhat remarkable, but in fact true that if the yo-yo does not slip it will wind itself up on the string.

 

[ubiquinone]

Hi, thank you for the help OlderDan.

Then I suppose for part d), once I've found the frictional force on the yo-yo, I can use the relationship, $$F_f=\mu F_N$$, where $$F_N=F_g$$ (the total weight of the yo-yo)?

 

[OlderDan]

Hi, thank you for the help OlderDan.

Then I suppose for part d), once I've found the frictional force on the yo-yo, I can use the relationship, $$F_f=\mu F_N$$, where $$F_N=F_g$$ (the total weight of the yo-yo)?

That will do it.