Concept of different accelerations on rotations of rigid bodies

[cyap]

Homework Statement

A cord passes over a massless and frictionless pulley, carrying a mass M1 at one end and wrapped around a cylinder of mass M2 which rolls on a horizontal plane. What is the acceleration of the mass M1?

Relevant Equations

τ=Iα
F=ma

I don't seem to understand the difference of accelerations when a cord is wrapped around cylinders and spheres. What I was taught is that if the cord is wrapped around a cylinder/sphere, the acceleration of the cord and whatever it is connected to will be twice the acceleration of the cylinder. Though I think it does not always apply, since for the question above I got $\frac{M_1g}{M_1+\frac{1}{4}M_2}$, which was not on the options

I'd like to know where the acceleration ratio comes from. Thanks in advance!

 

[Orodruin]

Can you describe how you got your result?

 

[kuruman]

What I was taught is that if the cord is wrapped around a cylinder/sphere, the acceleration of the cord and whatever it is connected to will be twice the acceleration of the cylinder.

You cannot apply that without scrutiny of how it was derived and where it applies especially what the "whatever" is. Assuming no slipping anywhere, it is true that the acceleration of the cord (and the hanging mass) is the same as at a point on the rim of each cylinder. Yet, the smaller cylinder has zero acceleration while the larger cylinder accelerates to the right.

 

[cyap]

Can you describe how you got your result?

$$Tr=(\frac{1}{2}M_2r^2)(\frac{a_2}{r})$$
$$T=\frac{1}{2}M_2a_2$$

$$M_1g-T=M_1a_1$$
$$M_1g-\frac{1}{2}M_2a_2=M_1a_1$$
Since according to "general knowledge" $a_2=\frac{1}{2}a_1$:
$$M_1g=M_1a_1+\frac{1}{4}M_2a_1$$
$$a_1=\frac{M_1g}{M_1+\frac{1}{4}M_2}$$

 

[cyap]

You cannot apply that without scrutiny of how it was derived and where it applies especially what the "whatever" is. Assuming no slipping anywhere, it is true that the acceleration of the cord (and the hanging mass) is the same as at a point on the rim of each cylinder. Yet, the smaller cylinder has zero acceleration while the larger cylinder accelerates to the right.

Which is exactly why I would love to know how it was derived

 

[kuruman]

Which is exactly why I would love to know how it was derived

You said that you were "taught" that. Who taught you that? Do you have any notes? Are you sure you remember correctly what you were taught?

Your derivation is incorrect. Look at the drawing on the right. It is a free body diagram (FBD) of the wheel.

• You wrote Newton's second law only for the angular acceleration about the center of mass and omitted Newton's second law for the linear acceleration of the center of mass, $a_{\text{cm}}$ which also involves the force of static friction $f_s$.
• The torque equation that you wrote, presumably about the center O, omits the torque due to static friction.

As for the "general knowledge" of the acceleration ratio, it can be deduced from the FBD. The wheel accelerates to the right without slipping which means that point P is instantaneously at rest. If the angular acceleration about point P is $\alpha$, the linear acceleration of point O is $a_O=a_{\text{cm}}=\alpha R.$ The linear acceleration of point Q on the rim is $a_Q=a_{\text{string}}=\alpha (2R).$ It follows that $a_{\text{string}}=2a_{\text{cm}}.$

You say that your answer was not one of the options. I pointed out where you went wrong. Can you post the options?

 

[Lnewqban]

I'd like to know where the acceleration ratio comes from. Thanks in advance!

The diameter equals two times the radius.

What would change regarding M1 and M2 velocities and accelerations if the cord is attached to the center of the cylinder, rather than be wrapped around it?

 

[cyap]

You said that you were "taught" that. Who taught you that? Do you have any notes? Are you sure you remember correctly what you were taught?

View attachment 350062Your derivation is incorrect. Look at the drawing on the right. It is a free body diagram (FBD) of the wheel.

• You wrote Newton's second law only for the angular acceleration about the center of mass and omitted Newton's second law for the linear acceleration of the center of mass, $a_{\text{cm}}$ which also involves the force of static friction $f_s$.
• The torque equation that you wrote, presumably about the center O, omits the torque due to static friction.

As for the "general knowledge" of the acceleration ratio, it can be deduced from the FBD. The wheel accelerates to the right without slipping which means that point P is instantaneously at rest. If the angular acceleration about point P is $\alpha$, the linear acceleration of point O is $a_O=a_{\text{cm}}=\alpha R.$ The linear acceleration of point Q on the rim is $a_Q=a_{\text{string}}=\alpha (2R).$ It follows that $a_{\text{string}}=2a_{\text{cm}}.$

You say that your answer was not one of the options. I pointed out where you went wrong. Can you post the options?

Wouldn't the friction be pointing to the right as the cylinder rotates clockwise?

 

[erobz]

Wouldn't the friction be pointing to the right as the cylinder rotates clockwise?

Perhaps you are being pushed to solve the equations of motion to verify. If the sign is reversed on $f$, it's the other way.

Put a coordinate system on the fixed pulley, and write a set equations using Newtons 2^nd ( Sum of Torques and Forces ) to solve said equations.

 

[erobz]

I used this convention:

So if you aren't sure which way friction acts, pick one. If you are sure, pick the right one. It won't be of consequence in the end result as long as the equations you develop are internally consistent with the conventions you chose.

 

[kuruman]

Wouldn't the friction be pointing to the right as the cylinder rotates clockwise?

In this case no. If you "turn off" friction by placing the cylinder on a frictionless surface, the cylinder will simply translate without rotating. If it hits a rough patch on the surface, the force of friction will have to be to the left in order to get the instantaneous point of contact to stop so that the cylinder rotates clockwise about it.

The argument above is incorrect. If the friction is turned off, the cylinder will still rotate about the center of mass because there will be a net torque about the CM. If you do the math correctly, you will see that the force of static friction will be in the same direction as the tension. Sorry about any confusion that my mental lapse may have created.