Chapter 16.1, Problem 16.44P

Disk B is at rest when it is brought into contact with disk A, which has an initial angular velocity ${\omega }_0$ . (a) Show that the final angular velocities of the disks are independent of the coefficient of friction ${\mu }_k$ between the disks as long as ${\mu }_k\ne 0$ . (b) Express the final angular velocity of disk A in terms of ${\omega }_0$ and the ratio of the masses of the two disks $m_A/m_B$ .

To determine

i.

Angular velocities have are independent of ${\text{μ}}_{\text{k}}$

Answer to Problem 16.44P

Hence proved, the final angular velocities have are independent of ${\text{μ}}_{\text{k}}$

Explanation of Solution

Given:

Disk A, of mass m_a radius of disk A, r_A

Initial Angular velocity of disk B = ${\text{ω}}_{\text{o}}$

mass of disk B, m_B

radius of disk B, r_B

Coefficient of kinetic friction = ${\text{μ}}_{\text{k}}$

Concept used:

When the two disks come in contact, a friction force between them comes into play and it causes the disk A to start rotating while accelerating with a certain angular acceleration in anti-clockwise direction. The reaction of the friction force on disk A will be acting on disk B, such that while it will still be rotating in clockwise direction, but with a certain. angular deacceleration. This will continue till the tangential velocity of both the disks become equal. At that point, ${\text{V}}_{\text{A}}{\text{ = V}}_{\text{B}}$

$\begin{array}{l}\text{V = rω,}\\ {\text{Therefore, r}}_{\text{A}}{\text{ω}}_{\text{A}}{\text{ = r}}_{\text{B}}{\text{ω}}_{\text{B}}\end{array}$

While accelerating from rest for disk A, ${\text{ω}}_{\text{A}}{\text{= ω}}_{\text{o}}{\text{ - α}}_{\text{A}}\text{t}$

While deaccelerating from an angular velocity ${\text{ω}}_{\text{o}}$ for disk B, ${\text{ω}}_B{\text{= 0 + α}}_B\text{t}$

Therefore, condition of velocity equivalence is

${\text{r}}_B{\text{( α}}_B{\text{t) = r}}_A\text{(}{\omega }_o{\text{-α}}_A\text{t) - (1)}$

Further, Mass moment of inertia for a disk is given by-

$\text{I = }\frac{1}{2}{\text{mr}}^2,\text{ where r is the radius of disk }$

The tangential force acting on a disk will provide the angular acceleration to the disk. Therefore,

$\begin{array}{l}\text{Summation of moments applied on it = Mass Moment of inertia }\times \text{ angular acceleration}\\ \Sigma \text{M = Iα -(2)}\end{array}$

Mass moment of inertia of disk A = $\frac{1}{2}{\text{m}}_{\text{A}}{r}_{\text{A}}^{\text{2}}$

Mass moment of inertia of disk B = $\frac{1}{2}{\text{m}}_B{r}_B^{\text{2}}$

Friction force between two disks, F = ${\text{μ}}_{\text{k}}{\text{N = μ}}_{\text{k}}{\text{m}}_{\text{B}}\text{g}$

For disk B,

From eq (2)

$\begin{array}{l}{\text{ΣM}}_B{\text{ = I}}_B{\text{α}}_B\\ {\text{F×r}}_B{\text{= I}}_B{\text{α}}_B\\ {\text{μ}}_{\text{k}}{\text{m}}_{\text{B}}{\text{g× r}}_B\text{= ( }\frac{1}{2}{\text{m}}_B{r}_B^{\text{2}}{\text{ ) × (α}}_B\text{)}\\ {\alpha }_B\text{= }\frac{2{\text{μ}}_{\text{k}}\text{g}}{r_B}{\text{ rad/s}}^{\text{2}}\end{array}$

For disk A,

From eq(2)

$\begin{array}{l}{\text{ΣM}}_{\text{A}}{\text{ = I}}_{\text{A}}{\text{α}}_{\text{A}}\\ {\text{μ}}_{\text{k}}{\text{m}}_{\text{B}}{\text{g× r}}_{\text{A}}\text{= }\frac{\text{1}}{\text{2}}{\text{m}}_{\text{A}}{\text{r}}_{\text{A}}^{\text{2}}{\text{ ×α}}_{\text{A}}\\ {\text{α}}_{\text{A}}\text{ = }\frac{{\text{2μ}}_{\text{k}}{\text{m}}_{\text{B}}\text{g}}{{\text{m}}_{\text{A}}{\text{r}}_{\text{A}}}{\text{ rad/s}}^{\text{2}}\end{array}$

Disk A will continue to deaccelerate, and disk B will continue to accelerate till their tangential acceleration becomes equal.

${\text{V}}_{\text{A}}{\text{ = V}}_{\text{B}}$

${\text{ r}}_{\text{A}}{\text{ω}}_{\text{A}}{\text{ = r}}_{\text{B}}{\text{ω}}_{\text{B}}$

from eq (1)

$\begin{array}{l}{\text{r}}_{\text{A}}{\text{(ω}}_{\text{o}}{\text{ - α}}_{\text{A}}{\text{t) = r}}_{\text{B}}{\text{(α}}_{\text{B}}\text{t)}\\ {\text{r}}_{\text{A}}{\text{(ω}}_{\text{o}}\text{ - }\frac{{\text{2μ}}_{\text{k}}{\text{m}}_{\text{B}}\text{g}}{{\text{m}}_{\text{A}}{\text{r}}_{\text{A}}}{\text{t) = r}}_{\text{B}}\text{(}\frac{{\text{2μ}}_{\text{k}}\text{g}}{{\text{r}}_{\text{B}}}\text{t)}\\ \text{t = }\frac{{\text{r}}_{\text{A}}{\text{ω}}_{\text{o}}}{{\text{2μ}}_{\text{k}}\text{g(}\frac{{\text{m}}_{\text{B}}}{{\text{m}}_{\text{A}}}\text{+1)}}\text{= }\frac{{\text{r}}_{\text{A}}{\text{ω}}_{\text{o}}{\text{m}}_{\text{A}}}{{\text{2μ}}_{\text{k}}{\text{g(m}}_{\text{B}}{\text{+m}}_{\text{A}}\text{)}}\end{array}$

$\begin{array}{l}{\text{ω}}_{\text{A}}{\text{ = ω}}_{\text{o}}{\text{ - α}}_{\text{A}}{\text{t = ω}}_{\text{o}}\text{ - (}\frac{{\text{2μ}}_{\text{k}}{\text{m}}_{\text{B}}\text{g}}{{\text{m}}_{\text{A}}{\text{r}}_{\text{A}}}\frac{{\text{r}}_{\text{A}}{\text{ω}}_{\text{o}}{\text{m}}_{\text{A}}}{{\text{2μ}}_{\text{k}}{\text{g(m}}_{\text{B}}{\text{+m}}_{\text{A}}\text{)}}{\text{) = ω}}_{\text{o}}{\text{ -(ω}}_{\text{o}}\frac{{\text{m}}_{\text{B}}}{{\text{m}}_{\text{B}}{\text{+m}}_{\text{A}}}\text{) -(3)}\\ {\text{ω}}_{\text{B}}{\text{ = α}}_{\text{B}}\text{t = }\frac{{\text{2μ}}_{\text{k}}\text{g}}{{\text{r}}_{\text{B}}}\frac{{\text{r}}_{\text{A}}{\text{ω}}_{\text{o}}{\text{m}}_{\text{A}}}{{\text{2μ}}_{\text{k}}{\text{g(m}}_{\text{B}}{\text{+m}}_{\text{A}}\text{)}}{\text{ = ω}}_{\text{o}}\frac{{\text{r}}_{\text{A}}}{{\text{r}}_{\text{B}}}\frac{{\text{m}}_{\text{A}}}{{\text{(m}}_{\text{B}}{\text{+m}}_{\text{A}}\text{)}}\text{ -(4)}\end{array}$

From equation (3) and (4), we see that the final angular velocities of the two disks are independent of ${\text{μ}}_{\text{k}}$, as long as ${\text{μ}}_{\text{k}}\ne {\text{ 0, because in that case, α}}_{\text{A}}{\text{=0,α}}_{\text{B}}{\text{=0, ω}}_{\text{A}}{\text{=ω}}_{\text{o}}{\text{, and ω}}_{\text{B}}\text{=0}$

Conclusion:

Hence proved, the final angular velocities have are independent of ${\text{μ}}_{\text{k}}$

To determine

ii.

Angular velocities in terms of in terms od ${\text{ω}}_{\text{o}}\text{ and }\frac{{\text{m}}_{\text{B}}}{{\text{m}}_{\text{A}}}$

Answer to Problem 16.44P

Angular velocity of disk is expressed as ${\text{ω}}_{\text{A}}{\text{= ω}}_{\text{o}}\text{(}\frac{\text{1}}{\text{1+}\frac{{\text{m}}_{\text{B}}}{{\text{m}}_{\text{A}}}}\text{) }$, in terms od ${\text{ω}}_{\text{o}}\text{ and }\frac{{\text{m}}_{\text{B}}}{{\text{m}}_{\text{A}}}$

Explanation of Solution

Given:

Disk A, of mass m_a radius of disk A, r_A

Initial Angular velocity of disk B = ${\text{ω}}_{\text{o}}$

mass of disk B, m_B

radius of disk B, r_B

Coefficient of kinetic friction = ${\text{μ}}_{\text{k}}$

Concept used:

When the two disks come in contact, a friction force between them comes into play and it causes the disk A to start rotating while accelerating with a certain angular acceleration in anti-clockwise direction. The reaction of the friction force on disk A will be acting on disk B, such that while it will still be rotating in clockwise direction, but with a certain. angular deacceleration. This will continue till the tangential velocity of both the disks become equal. At that point, ${\text{V}}_{\text{A}}{\text{ = V}}_{\text{B}}$

$\begin{array}{l}\text{V = rω,}\\ {\text{Therefore, r}}_{\text{A}}{\text{ω}}_{\text{A}}{\text{ = r}}_{\text{B}}{\text{ω}}_{\text{B}}\end{array}$

While accelerating from rest for disk A, ${\text{ω}}_{\text{A}}{\text{= ω}}_{\text{o}}{\text{ - α}}_{\text{A}}\text{t}$

While deaccelerating from an angular velocity ${\text{ω}}_{\text{o}}$ for disk B, ${\text{ω}}_B{\text{= 0 + α}}_B\text{t}$

Therefore, condition of velocity equivalence is

${\text{r}}_B{\text{( α}}_B{\text{t) = r}}_A\text{(}{\omega }_o{\text{-α}}_A\text{t) - (1)}$

Further, Mass moment of inertia for a disk is given by-

$\text{I = }\frac{1}{2}{\text{mr}}^2,\text{ where r is the radius of disk }$

The tangential force acting on a disk will provide the angular acceleration to the disk. Therefore,

$\begin{array}{l}\text{Summation of moments applied on it = Mass Moment of inertia }\times \text{ angular acceleration}\\ \Sigma \text{M = Iα -(2)}\end{array}$

Mass moment of inertia of disk A = $\frac{1}{2}{\text{m}}_{\text{A}}{r}_{\text{A}}^{\text{2}}$

Mass moment of inertia of disk B = $\frac{1}{2}{\text{m}}_B{r}_B^{\text{2}}$

Friction force between two disks, F = ${\text{μ}}_{\text{k}}{\text{N = μ}}_{\text{k}}{\text{m}}_{\text{B}}\text{g}$

For disk B,

From eq (2)

$\begin{array}{l}{\text{ΣM}}_B{\text{ = I}}_B{\text{α}}_B\\ {\text{F×r}}_B{\text{= I}}_B{\text{α}}_B\\ {\text{μ}}_{\text{k}}{\text{m}}_{\text{B}}{\text{g× r}}_B\text{= ( }\frac{1}{2}{\text{m}}_B{r}_B^{\text{2}}{\text{ ) × (α}}_B\text{)}\\ {\alpha }_B\text{= }\frac{2{\text{μ}}_{\text{k}}\text{g}}{r_B}{\text{ rad/s}}^{\text{2}}\end{array}$

For disk A,

From eq(2)

$\begin{array}{l}{\text{ΣM}}_{\text{A}}{\text{ = I}}_{\text{A}}{\text{α}}_{\text{A}}\\ {\text{μ}}_{\text{k}}{\text{m}}_{\text{B}}{\text{g× r}}_{\text{A}}\text{= }\frac{\text{1}}{\text{2}}{\text{m}}_{\text{A}}{\text{r}}_{\text{A}}^{\text{2}}{\text{ ×α}}_{\text{A}}\\ {\text{α}}_{\text{A}}\text{ = }\frac{{\text{2μ}}_{\text{k}}{\text{m}}_{\text{B}}\text{g}}{{\text{m}}_{\text{A}}{\text{r}}_{\text{A}}}{\text{ rad/s}}^{\text{2}}\end{array}$

Disk A will continue to deaccelerate, and disk B will continue to accelerate till their tangential acceleration becomes equal.

${\text{V}}_{\text{A}}{\text{ = V}}_{\text{B}}$

from eq (1)

$\begin{array}{l}{\text{r}}_{\text{A}}{\text{(ω}}_{\text{o}}{\text{ - α}}_{\text{A}}{\text{t) = r}}_{\text{B}}{\text{(α}}_{\text{B}}\text{t)}\\ {\text{r}}_{\text{A}}{\text{(ω}}_{\text{o}}\text{ - }\frac{{\text{2μ}}_{\text{k}}{\text{m}}_{\text{B}}\text{g}}{{\text{m}}_{\text{A}}{\text{r}}_{\text{A}}}{\text{t) = r}}_{\text{B}}\text{(}\frac{{\text{2μ}}_{\text{k}}\text{g}}{{\text{r}}_{\text{B}}}\text{t)}\\ \text{t = }\frac{{\text{r}}_{\text{A}}{\text{ω}}_{\text{o}}}{{\text{2μ}}_{\text{k}}\text{g(}\frac{{\text{m}}_{\text{B}}}{{\text{m}}_{\text{A}}}\text{+1)}}\text{= }\frac{{\text{r}}_{\text{A}}{\text{ω}}_{\text{o}}{\text{m}}_{\text{A}}}{{\text{2μ}}_{\text{k}}{\text{g(m}}_{\text{B}}{\text{+m}}_{\text{A}}\text{)}}\end{array}$

$\begin{array}{l}{\text{ω}}_{\text{A}}{\text{ = ω}}_{\text{o}}{\text{ - α}}_{\text{A}}{\text{t = ω}}_{\text{o}}\text{ - (}\frac{{\text{2μ}}_{\text{k}}{\text{m}}_{\text{B}}\text{g}}{{\text{m}}_{\text{A}}{\text{r}}_{\text{A}}}\frac{{\text{r}}_{\text{A}}{\text{ω}}_{\text{o}}{\text{m}}_{\text{A}}}{{\text{2μ}}_{\text{k}}{\text{g(m}}_{\text{B}}{\text{+m}}_{\text{A}}\text{)}}{\text{) = ω}}_{\text{o}}{\text{ -(ω}}_{\text{o}}\frac{{\text{m}}_{\text{B}}}{{\text{m}}_{\text{B}}{\text{+m}}_{\text{A}}}\text{) -(3)}\\ {\text{ω}}_{\text{B}}{\text{ = α}}_{\text{B}}\text{t = }\frac{{\text{2μ}}_{\text{k}}\text{g}}{{\text{r}}_{\text{B}}}\frac{{\text{r}}_{\text{A}}{\text{ω}}_{\text{o}}{\text{m}}_{\text{A}}}{{\text{2μ}}_{\text{k}}{\text{g(m}}_{\text{B}}{\text{+m}}_{\text{A}}\text{)}}{\text{ = ω}}_{\text{o}}\frac{{\text{r}}_{\text{A}}}{{\text{r}}_{\text{B}}}\frac{{\text{m}}_{\text{A}}}{{\text{(m}}_{\text{B}}{\text{+m}}_{\text{A}}\text{)}}\text{ -(4)}\\ {\text{ω}}_{\text{A}}{\text{= ω}}_{\text{o}}\text{ (1-(}\frac{{\text{m}}_{\text{B}}}{{\text{m}}_{\text{B}}{\text{+m}}_{\text{A}}}{\text{)) = ω}}_{\text{o}}\text{(}\frac{{\text{m}}_{\text{B}}{\text{+m}}_{\text{A}}{\text{-m}}_{\text{B}}}{{\text{m}}_{\text{B}}{\text{+m}}_{\text{A}}}{\text{) = ω}}_{\text{o}}\text{(}\frac{{\text{m}}_{\text{A}}}{{\text{m}}_{\text{B}}{\text{+m}}_{\text{A}}}\text{)}\\ {\text{ω}}_{\text{A}}{\text{= ω}}_{\text{o}}\text{(}\frac{\text{1}}{\text{1+}\frac{{\text{m}}_{\text{B}}}{{\text{m}}_{\text{A}}}}\text{) -(5)}\end{array}$

Conclusion:

Angular velocity of disk i

Is expressed as ${\text{ω}}_{\text{A}}{\text{= ω}}_{\text{o}}\text{(}\frac{\text{1}}{\text{1+}\frac{{\text{m}}_{\text{B}}}{{\text{m}}_{\text{A}}}}\text{) }$, in terms od ${\text{ω}}_{\text{o}}\text{ and }\frac{{\text{m}}_{\text{B}}}{{\text{m}}_{\text{A}}}$