Circular motion of a mass on a string on an inclined plane

[ThEmptyTree]

Homework Statement

A body of mass $m$ is attached to one end of a string of length $R$. The other end of the string is fixed on an inclined plane making an angle $\phi$ with the horizontal as shown in the figure. The body has speed $v_0$ at the bottom of the circle (point $A$). The body undergoes circular motion. There is a coefficient of sliding friction $\mu$ between the body and the plane. The downward acceleration of gravity is $g$. Express all answers in terms of $m, \phi, v_0, g, \mu$ and $R$ as needed.

(a) How much work does the friction force do on the body as it moves from the bottom of the circle (point $A$) to the top of the circle (point $B$)?
(b) What is the tension in the string when it reaches point $B$? Express your answer in terms of $m, \phi, v_0, g, \mu$ and $R$ as needed.

Relevant Equations

Work-Kinetic Energy theorem:
$$\Delta{K}=W$$

(I drew motion in the opposite direction so the object would rotate trigonometrically but it should be the same thing)

I have just finished the Kinetic Energy and Work chapter in my course and this is the last problem from the problem set. I have not worked many problems with the Work-Kinetic Energy Theorem on non-linear paths and I would like to know whether I am doing well or not.

(a) Definition of work: $$(W_f)_{A,B}=\int\limits_{\vec{r}=\vec{r_A}}^{\vec{r}=\vec{r_B}} \vec{f}\cdot d\vec{r}$$
Projection of vectors :
$$\vec{f}=-f\hat{\theta}=-\mu N \hat{\theta}=-\mu mg cos\phi \hat{\theta}$$
$$d\vec{r}=Rd\theta \hat{\theta}$$
Scalar product: $$\vec{f}\cdot d\vec{r}=-\mu mg R cos\phi d\theta$$
Evaluating the integral: $$(W_f)_{A,B}=\int\limits_{\theta=0}^{\theta=\pi} -\mu mg R cos\phi d\theta=-\mu mg \pi R cos\phi$$
(b) Work-Kinetic Energy Theorem: $$\Delta{K_{A,B}}=K_B-K_A=\frac{1}{2}m(v_f^2-v_0^2)$$
Total work: $$W_{A,B}=\underbrace{(W_T)_{A,B}}_{0}+(W_f)_{A,B}=-\mu mg \pi R cos\phi$$
Equating: $$v_f^2=v_0^2-2\mu g \pi R cos\phi$$
Radial acceleration for circular motion: $$a_r=\frac{v^2}{R}$$
Newton's 2nd law to evaluate radial forces in $B$: $$-T_B=ma_r\Rightarrow T_B=m\big(\frac{v_0^2}{R}-2\mu g \pi cos\phi\big)$$

The answer at (a) is intuitive because $-\mu mg cos\phi$ is the force acting and $\pi R$ is half the circle as if the object were to move in a straight line. However I have no idea if (b) is correct. Please help! uwu

 

[haruspex]

if (b) is correct

No.
You know what the acceleration is at B, both direction and magnitude.
What forces contribute to that acceleration?

 

[ThEmptyTree]

No.
You know what the acceleration is at B, both direction and magnitude.
What forces contribute to that acceleration?

Damn, forgot the gravity. Knew it was too easy.

Though I think I know how to take care of gravity, it is a conservative force and its expression it pretty simple. I just have to add the work done by gravity to total work
$$\Delta{y}_{A,B}=2 R sin\phi$$
$$(W_{grav})_{A,B}=-mg\Delta{y}_{A,B}=-2 mg R sin\phi$$
and project it on $\hat{r_B}$
$$-T_B-mg cos\phi=ma_r$$

The only other force I can think of is the normal, but being perpendicular to the plane its work is null and it doesn't project on $\hat{r_B}$ so I just ignored it.

Is there anything else I am missing?

 

[haruspex]

I'm not seeing enough detail above to say whether you have it now.
What is your final answer in full?

 

[ThEmptyTree]

I'm not seeing enough detail above to say whether you have it now.
What is your final answer in full?

I will remake (b).

Change in kinetic energy : $$\Delta{K}_{A,B}=K_B-K_A=\frac{1}{2}m(v_f^2-v_0^2)$$
Vertical displacement : $$\Delta{y}_{A,B}=2 R sin\phi$$
Work done by gravity : $$(W_{grav})_{A,B}=-mg\Delta{y}_{A,B}=-2 mg R sin\phi$$
Total work : $$W_{A,B}=(W_{grav})_{A.B}+(W_f)_{A.B}+\underbrace{(W_N)_{A,B}}_{0}+\underbrace{(W_T)_{A,B}}_{0}=-mgR(2sin\phi + \mu \pi cos\phi)$$
Equating through the "work-kinetic energy theorem" : $$v_f^2=v_0^2-2gR(2sin\phi + \mu \pi cos\phi)$$
Net force and acceleration on $\hat{r_B}$ :
$$\Sigma F=-T_B-mg cos\phi$$
$$m a_r=-\frac{mv_f^2}{R}=-m\big(\frac{v_0^2}{R}-2g(2sin\phi+\mu \pi cos\phi)\big)$$
Equating through Newton's 2nd law : $$T_B=m\Big(\frac{v_0^2}{R}-g\big(4sin\phi+(1 + 2 \mu \pi)cos\phi\big)\Big)$$

 

[haruspex]

Shouldn't the string loosen up tension as the object gets closer to the top?

You have a couple of sign errors. For a start, speed should reduce.

 

[ThEmptyTree]

You have a couple of sign errors. For a start, speed should reduce.

It is so satisfying transforming a bunch of + into - signs on paper just by erasing the little lines above and under the -.

I have edited. Is it good now?

 

[haruspex]

It is so satisfying transforming a bunch of + into - signs on paper just by erasing the little lines above and under the -.

I have edited. Is it good now?

Not quite. Track the $mg\cos(\phi)$ term through, watching the sign.

 

[ThEmptyTree]

You are right, I have corrected it.

I have done these late night and taking into account that I haven't slept I think I am so mentally disabled right now making these errors. Excuse my inattention and impatience.

 

[haruspex]

You are right, I have corrected it.

I have done these late night and taking into account that I haven't slept I think I am so mentally disabled right now making these errors. Excuse my inattention and impatience.

Looks good, now get some sleep!

 

[ThEmptyTree]

Thanks for the help!

 

[StoicHacker]

I will remake (b).

Change in kinetic energy : $$\Delta{K}_{A,B}=K_B-K_A=\frac{1}{2}m(v_f^2-v_0^2)$$
Vertical displacement : $$\Delta{y}_{A,B}=2 R sin\phi$$
Work done by gravity : $$(W_{grav})_{A,B}=-mg\Delta{y}_{A,B}=-2 mg R sin\phi$$
Total work : $$W_{A,B}=(W_{grav})_{A.B}+(W_f)_{A.B}+\underbrace{(W_N)_{A,B}}_{0}+\underbrace{(W_T)_{A,B}}_{0}=-mgR(2sin\phi + \mu \pi cos\phi)$$
Equating through the "work-kinetic energy theorem" : $$v_f^2=v_0^2-2gR(2sin\phi + \mu \pi cos\phi)$$
Net force and acceleration on $\hat{r_B}$ :
$$\Sigma F=-T_B-mg cos\phi$$
$$m a_r=-\frac{mv_f^2}{R}=-m\big(\frac{v_0^2}{R}-2g(2sin\phi+\mu \pi cos\phi)\big)$$
Equating through Newton's 2nd law : $$T_B=m\Big(\frac{v_0^2}{R}-g\big(4sin\phi+(1 + 2 \mu \pi)cos\phi\big)\Big)$$

I am not sure how did you get the component of the weight in the radial direction towards the center at point b to involve a cosine. It should be a sine instead. That would obviously change the final solution but other than that everything else is good.

 

[kuruman]

I am not sure how did you get the component of the weight in the radial direction towards the center at point b to involve a cosine.

The term $mg\cos\phi$ in the solution is the component of the weight perpendicular to the plane, i.e. the normal force. It is used to find the force of kinetic friction and hence the work done by kinetic friction, $W_f=-\mu (mg\cos\phi)(\pi R)$. It is not meant to be the component of the weight in the radial direction.

 

[StoicHacker]

The term $mg\cos\phi$ in the solution is the component of the weight perpendicular to the plane, i.e. the normal force. It is used to find the force of kinetic friction and hence the work done by kinetic friction, $W_f=-\mu (mg\cos\phi)(\pi R)$. It is not meant to be the component of the weight in the radial direction.

I was referring to using cosine in Newton's law. This:
$$ \sum F = - T_{B} - mgcos(\phi)$$

Should be:

$$ \sum F = - T_{B} - mgsin(\phi)$$

It is the component of the weight parallel to the plane that contributes to the radial acceleration and not the perpendicular component, i.e the normal.

 

[kuruman]

I was referring to using cosine in Newton's law. This:
$$ \sum F = - T_{B} - mgcos(\phi)$$

Should be:

$$ \sum F = - T_{B} - mgsin(\phi)$$

It is the component of the weight parallel to the plane that contributes to the radial acceleration and not the perpendicular component, i.e the normal.

Yes, that is correct. The final expression should be
$$T_B=m\Big[\frac{v_0^2}{R}-g\big(5\sin\phi+2 \mu \pi\cos\phi\big)\Big].$$