Confusion regarding signs in rotational motion

[Eggue]

Homework Statement

A solid cylinder with mass M and radius R, rotating with angular speed about an axis through its center, is set on a horizontal surface for which the kinetic friction coefficient is (a) Draw a free-body diagram for the cylinder on the surface. Think carefully about the direction of the kinetic friction force on the cylinder. Calculate the accelerations of the center of mass and of rotation about the center of mass. (b) The cylinder is initially slipping completely, so initially but Rolling without slipping sets in when Calculate the distance the cylinder rolls before slipping stops

Relevant Equations

Kinematic equations

I'm not sure as to why my working is incorrect. When the sign on a_x is postive, i get t = \frac{R\omega_0}{3\mu_kg} which would give the correct value for distance if plugged into the kinematic equation. However, I'm not sure why a_x would be positive though since the friction force is pointing in the direction of -x

Working:

 

[jbriggs444]

I'm not sure as to why my working is incorrect. When the sign on a_x is postive, i get t = \frac{R\omega_0}{3\mu_kg} which would give the correct value for distance if plugged into the kinematic equation. However, I'm not sure why a_x would be positive though since the friction force is pointing in the direction of -x

In your hand-drawn solving, you write an equation:$$-d_f=d_0 + v_0t + \frac{1}{2}a_xt^2$$Can you justify that equation? Why the minus sign on $d_f$?

 

[wrobel]

There is a top secret method of how never get confused with signs. First write equations in vector form then expand them in coordinate frame

 

[Eggue]

In your hand-drawn solving, you write an equation:$$-d_f=d_0 + v_0t + \frac{1}{2}a_xt^2$$Can you justify that equation? Why the minus sign on $d_f$?

Oh that was a mistake that i corrected later, but that shouldn't affect anything right?

 

[jbriggs444]

Oh that was a mistake that i corrected later, but that shouldn't affect anything right?

It affects the sign of $d_f$ surely. I thought that was what you were asking about -- the sign on the value for $d_f$.

 

[Eggue]

It affects the sign of $d_f$ surely. I thought that was what you were asking about -- the sign on $d_f$.

No what i meant was the sign on a_x changes the value of t and i do not know why the sign should be positive even though friction is pointing in the -x direction.

 

[jbriggs444]

Well, let us review how you are calculating t. It seems that we will have to decode some chicken-scratchings.

You start with$$F_k=\mu_kN=-\mu_kmg$$At this point you have inserted a sign into the formula because the relative motion of the ground beneath the cylinder is leftward -- in the direction of negative x. I agree. Because the only force acting is $F_k$, you invoke Newton's second law ($F=ma$) and assert that$$ma_x=-\mu_kmg$$. A straightforward inference. I agree. You divide out the m to obtain$$a_x=-\mu_kg$$You now shift to angular momentum and assert that equate the rate of change of angular momentum with the applied torque:$$I\alpha_z=F_kR$$Here you are implicitly adopting a sign convention such that $\alpha_z$ is positive counter-clockwise. You proceed to substitute in for the moment of inertia $I=\frac{1}{2}MR^2$ and $F_k=-\mu_kmg$ to obtain$$\frac{1}{2}mR^2\alpha_z=-\mu_kmgR$$You are a bit sloppy with the case on the M versus m, but you cancel it anyway along with one factor of R to obtain$$\frac{1}{2}R\alpha_z=-\mu_kmg$$You proceed to solve for $\alpha_z$ obtaining$$\alpha_z=\frac{-2\mu_kg}{R}$$I agree with this and agree that the sign on the resulting angular acceleration will be negative (i.e. clockwise).

You now do some stuff with variable names, but since you never actually use any of the following variable names, it turns out not to matter.

You introduce $\omega_z$ equal to the starting rotation rate.

You introduce $v_x$ equal to zero.

You introduce $v_{x_1}$ equal to the rim speed relative to the center of the cylinder.

The equation that matters is $$v_0+a_xt =R(\omega_0+\alpha_zt)$$This is supposed to be the rolling without slipping equation. But let us examine it for sign convention agreement. On the left hand side we have the velocity of the cylinder with a sign convention of right = positive. On the right hand side we have the rim speed of the cylinder with a sign convention of counter-clockwise positive. But that does not fit. A positive (rightward) velocity without slipping means a negative (clockwise) roll rate.

There's your sign error.

 

[Eggue]

Well, let us review how you are calculating t. It seems that we will have to decode some chicken-scratchings.

You start with$$F_k=\mu_kN=-\mu_kmg$$At this point you have inserted a sign into the formula because the relative motion of the ground beneath the cylinder is leftward -- in the direction of negative x. I agree. Because the only force acting is $F_k$, you invoke Newton's second law ($F=ma$) and assert that$$ma_x=-\mu_kmg$$. A straightforward inference. I agree. You divide out the m to obtain$$a_x=-\mu_kg$$You now shift to angular momentum and assert that equate the rate of change of angular momentum with the applied torque:$$I\alpha_z=F_kR$$Here you are implicitly adopting a sign convention such that $\alpha_z$ is positive counter-clockwise. You proceed to substitute in for the moment of inertia $I=\frac{1}{2}MR^2$ and $F_k=-\mu_kmg$ to obtain$$\frac{1}{2}mR^2\alpha_z=-\mu_kmgR$$You are a bit sloppy with the case on the M versus m, but you cancel it anyway along with one factor of R to obtain$$\frac{1}{2}R\alpha_z=-\mu_kmg$$You proceed to solve for $\alpha_z$ obtaining$$\alpha_z=\frac{-2\mu_kg}{R}$$I agree with this and agree that the sign on the resulting angular acceleration will be negative (i.e. clockwise).

You now do some stuff with variable names, but since you never actually use any of the following variable names, it turns out not to matter.

You introduce $\omega_z$ equal to the starting rotation rate.

You introduce $v_x$ equal to zero.

You introduce $v_{x_1}$ equal to the rim speed relative to the center of the cylinder.

The equation that matters is $$v_0+a_xt =R(\omega_0+\alpha_zt)$$This is supposed to be the rolling without slipping equation. But let us examine it for sign convention agreement. On the left hand side we have the velocity of the cylinder with a sign convention of right = positive. On the right hand side we have the rim speed of the cylinder with a sign convention of counter-clockwise positive. But that does not fit. A positive (rightward) velocity without slipping means a negative (clockwise) roll rate.

There's your sign error.

Ah i understand now. Thank you