Return to Question

A small disc is projected on a horizontal floor with speed u. Coefficient of friction between the disc and floor varies as $\mu = \mu_0+ kx$, where $x$ is the distance covered. Find distance slid by the disc on the floor.

I have correctly solved this by using two methods :

1. By integrating the kinematic equation $\displaystyle a=v\frac{dv}{dx}=-(\mu_0+kx)g$ and
2. by considering the work done by frictional force ($\int_0^x (\mu_0+kx)mgdx$) to dissipate the initial kinetic energy of the disc.

Now I notice that the equation of motion, $a=-(\mu_0 +kx)g$, is kinda SHM-like, but with a non-conservative force instead. I conjecture that as soon as the disc reaches $v=0$, the force of friction, and hence the acceleration, drop to 0. Now, I think that this is pretty similar (except the $a=0$ at $v=0$ part; in SHM, from my knowledge, $a$ and $v$ are separated by a phase angle of $\frac π2$) to the first $\frac 14$ of an SHM cycle, i.e. starting from the mean position till it reaches the amplitude.

Now my calculations for the third method: The maximum speed obtained is at mean position: $v_{max}=u=A\omega$ where $\omega$ is the angular frequency of oscillation. Also, $\omega ^2=kg$ and so maximum distance travelled is equal to the amplitude, i.e, when velocity becomes 0. So $$ A=\frac{u}{\sqrt{kg}}. $$ However, this is not the correct answer.

What’s my error in concept?

EDIT: I understood where I got confused, thanks to user @J.Murray. But I’d like to see this question solved (or get the easily solvable equation $kgx^2+2\mu_0gx-u^2=0$) with concepts borrowed from SHM, and not directly the two methods I have listed above. No big deal if it is complicated.

A small disc is projected on a horizontal floor with speed $u$. Coefficient of friction between the disc and floor varies as $\mu = \mu_0+ kx$, where $x$ is the distance covered. Find distance slid by the disc on the floor.

I have correctly solved this by using two methods :

1. By integrating the kinematic equation $\displaystyle a=v\frac{dv}{dx}=-(\mu_0+kx)g$ and
2. by considering the work done by frictional force ($\int_0^x (\mu_0+kx)mgdx$) to dissipate the initial kinetic energy of the disc.

Now I notice that the equation of motion, $a=-(\mu_0 +kx)g$, is kinda SHM-like, but with a non-conservative force instead. I conjecture that as soon as the disc reaches $v=0$, the force of friction, and hence the acceleration, drop to 0. Now, I think that this is pretty similar (except the $a=0$ at $v=0$ part; in SHM, from my knowledge, $a$ and $v$ are separated by a phase angle of $\frac π2$) to the first $\frac 14$ of an SHM cycle, i.e. starting from the mean position till it reaches the amplitude.

Now my calculations for the third method: The maximum speed obtained is at mean position: $v_{max}=u=A\omega$ where $\omega$ is the angular frequency of oscillation. Also, $\omega ^2=kg$ and so maximum distance travelled is equal to the amplitude, i.e, when velocity becomes 0. So $$ A=\frac{u}{\sqrt{kg}}. $$ However, this is not the correct answer.

What’s my error in concept?

EDIT: I understood where I got confused, thanks to user @J.Murray. But I’d like to see this question solved (or get the easily solvable equation $kgx^2+2\mu_0gx-u^2=0$) with concepts borrowed from SHM, and not directly the two methods I have listed above. No big deal if it is complicated.

A small disc is projected on a horizontal floor with speed u. Coefficient of friction between the disc and floor varies as $\mu = \mu_0+ kx$, where $x$ is the distance covered. Find distance slid by the disc on the floor.

I have correctly solved this by using two methods :

1. By integrating the kinematic equation $\displaystyle a=v\frac{dv}{dx}=-(\mu_0+kx)g$ and
2. by considering the work done by frictional force ($\int_0^x (\mu_0+kx)mgdx$) to dissipate the initial kinetic energy of the disc.

Now I notice that the equation of motion, $a=-(\mu_0 +kx)g$, is kinda SHM-like, but with a non-conservative force instead. I conjecture that as soon as the disc reaches $v=0$, the force of friction, and hence the acceleration, drop to 0. Now, I think that this is pretty similar (except the $a=0$ at $v=0$ part; in SHM, from my knowledge, $a$ and $v$ are separated by a phase angle of $\frac π2$) to the first $\frac 14$ of an SHM cycle, i.e. starting from the mean position till it reaches the amplitude.

Now my calculations for the third method: The maximum speed obtained is at mean position: $v_{max}=u=A\omega$ where $\omega$ is the angular frequency of oscillation. Also, $\omega ^2=kg$ and so maximum distance travelled is equal to the amplitude, i.e, when velocity becomes 0. So $$ A=\frac{u}{\sqrt{kg}}. $$ However, this is not the correct answer.

What’s my error in concept?

A small disc is projected on a horizontal floor with speed u. Coefficient of friction between the disc and floor varies as $\mu = \mu_0+ kx$, where $x$ is the distance covered. Find distance slid by the disc on the floor.

I have correctly solved this by using two methods :

1. By integrating the kinematic equation $\displaystyle a=v\frac{dv}{dx}=-(\mu_0+kx)g$ and
2. by considering the work done by frictional force ($\int_0^x (\mu_0+kx)mgdx$) to dissipate the initial kinetic energy of the disc.

Now I notice that the equation of motion, $a=-(\mu_0 +kx)g$, is kinda SHM-like, but with a non-conservative force instead. I conjecture that as soon as the disc reaches $v=0$, the force of friction, and hence the acceleration, drop to 0. Now, I think that this is pretty similar (except the $a=0$ at $v=0$ part; in SHM, from my knowledge, $a$ and $v$ are separated by a phase angle of $\frac π2$) to the first $\frac 14$ of an SHM cycle, i.e. starting from the mean position till it reaches the amplitude.

Now my calculations for the third method: The maximum speed obtained is at mean position: $v_{max}=u=A\omega$ where $\omega$ is the angular frequency of oscillation. Also, $\omega ^2=kg$ and so maximum distance travelled is equal to the amplitude, i.e, when velocity becomes 0. So $$ A=\frac{u}{\sqrt{kg}}. $$ However, this is not the correct answer.

What’s my error in concept?

EDIT: I understood where I got confused, thanks to user @J.Murray. But I’d like to see this question solved (or get the easily solvable equation $kgx^2+2\mu_0gx-u^2=0$) with concepts borrowed from SHM, and not directly the two methods I have listed above. No big deal if it is complicated.

A small disc is projected on a horizontal floor with speed u. Coefficient of friction between the disc and floor varies as $\mu = \mu_0+ kx$, where $x$ is the distance covered. Find distance slid by the disc on the floor.

I have correctly solved this by using two methods :

1. By integrating the kinematic equation $\displaystyle a=v\frac{dv}{dx}=-(\mu_0+kx)g$ and
2. by considering the work done by frictional force ($\int_0^x m(\mu_0+kx)gdx$) to dissipate the initial kinetic energy of the disc.

Now I notice that the equation of motion, $a=-(\mu_0 +kx)g$, is kinda SHM-like, but with a non-conservative force instead. I conjecture that as soon as the disc reaches $v=0$, the force of friction, and hence the acceleration, drop to 0. Now, I think that this is pretty similar (except the $a=0$ at $v=0$ part; in SHM, from my knowledge, $a$ and $v$ are separated by a phase angle of $\frac π2$) to the first $\frac 14$ of an SHM cycle, i.e. starting from the mean position till it reaches the amplitude.

Now my calculations for the third method: The maximum speed obtained is at mean position: $v_{max}=u=A\omega$ where $\omega$ is the angular frequency of oscillation. Also, $\omega ^2=kg$ and so maximum distance travelled is equal to the amplitude, i.e, when velocity becomes 0. So $$ A=\frac{u}{\sqrt{kg}}. $$ However, this is not the correct answer.

What’s my error in concept?

A small disc is projected on a horizontal floor with speed u. Coefficient of friction between the disc and floor varies as $\mu = \mu_0+ kx$, where $x$ is the distance covered. Find distance slid by the disc on the floor.

I have correctly solved this by using two methods :

1. By integrating the kinematic equation $\displaystyle a=v\frac{dv}{dx}=-(\mu_0+kx)g$ and
2. by considering the work done by frictional force ($\int_0^x (\mu_0+kx)mgdx$) to dissipate the initial kinetic energy of the disc.

Now I notice that the equation of motion, $a=-(\mu_0 +kx)g$, is kinda SHM-like, but with a non-conservative force instead. I conjecture that as soon as the disc reaches $v=0$, the force of friction, and hence the acceleration, drop to 0. Now, I think that this is pretty similar (except the $a=0$ at $v=0$ part; in SHM, from my knowledge, $a$ and $v$ are separated by a phase angle of $\frac π2$) to the first $\frac 14$ of an SHM cycle, i.e. starting from the mean position till it reaches the amplitude.

Now my calculations for the third method: The maximum speed obtained is at mean position: $v_{max}=u=A\omega$ where $\omega$ is the angular frequency of oscillation. Also, $\omega ^2=kg$ and so maximum distance travelled is equal to the amplitude, i.e, when velocity becomes 0. So $$ A=\frac{u}{\sqrt{kg}}. $$ However, this is not the correct answer.

What’s my error in concept?