Chapter 16.1, Problem 16.74P

A sphere of radius r and mass m has a linear velocity v₀ directed to the left and no angular velocity as it is placed on a belt moving to the right with a constant velocity v₁. If after first sliding on the belt the sphere is to have no linear velocity relative to the ground as it starts rolling on the belt without sliding, determine in terms of v₁and the coefficient of kinetic friction ${\mu }_k$ between the sphere and the belt (a) the required value of v₀, (b) the time t₁at which the sphere will start rolling on the belt, (c) the distance the sphere will have moved relative to the ground at time t₁.

To determine

(a)

The value of $v_0$.

Answer to Problem 16.74P

Value of $v_0=\frac{2v_1}{5}$

Explanation of Solution

Given information:

Mass $m$

Radius $r$

Belt velocity $v_1$

Ball velocity $v_0$

Friction coefficient ${\mu }_k$

Concept used:

Following formula is used-

1. Sum of horizontal forces, $\sum F_x=ma$

2. Sum of moments about mass center, $\sum M_G=I\alpha$

Calculation:

Friction force:

$\begin{array}{l}f={\mu }_{k}N\\ f={\mu }_{k}mg\end{array}$

Sum of horizontal forces;

$\begin{array}{l}\sum F_x=ma\\ f=ma\\ {\mu }_{k}mg=ma\\ a={\mu }_{k}g\to \end{array}$

Sum of moments about mass center:

$\begin{array}{l}\sum M_G=I\alpha \\ f\times r=m{k}^2\alpha \\ {\mu }_{k}mg\times r=m{k}^2\alpha \text{ for sphere }{k}^2=\frac{2}{5}{r}^2\\ \alpha =\frac{5{\mu }_{k}g}{2r}{\text{rad/s}}^{\text{2}}\end{array}$

Velocity equation,

$\begin{array}{l}{v}_{}=v_0-at\leftarrow \\ v_{}=v_0-{\mu }_{k}gt\leftarrow \end{array}$

Angular velocity equation,

$\begin{array}{l}\omega ={\omega }_0-\alpha t\\ \omega =\frac{5{\mu }_{k}g}{2r}t\end{array}$

Velocity of contact point,

$\begin{array}{l}{v}_C=-v+r\omega \\ v_C=-v+r(\frac{5{\mu }_{k}g}{2r}t)\\ v_C=-v+(\frac{5{\mu }_{k}g}{2}t)\end{array}$

$\begin{array}{l}{\text{when t=t}}_{\text{1}},{\text{ v=0 so v}}_{\text{c}}{\text{=v}}_{\text{1}}\\ v_1=(\frac{5{\mu }_{k}g{t}_1}{2})\\ t_1=\frac{2v_1}{5{\mu }_{k}g}\end{array}$

$\begin{array}{l}{v}_{}=v_0-{\mu }_{k}gt\\ 0=v_0-{\mu }_{k}g\frac{2v_1}{5{\mu }_{k}g}\\ v_0=\frac{2v_1}{5}\end{array}$

Conclusion:

Thus we get,

Value of $v_0=\frac{2v_1}{5}$

To determine

(b)

The time at which sphere starts rolling.

Answer to Problem 16.74P

Value of time $t_1=\frac{2v_1}{5{\mu }_{k}g}$

Explanation of Solution

Given information:

Mass $m$

Radius $r$

Belt velocity $v_1$

Ball velocity $v_0$

Friction coefficient ${\mu }_k$

Concept used:

Following formula is used-

1. Sum of horizontal forces, $\sum F_x=ma$

2. Sum of moments about mass center, $\sum M_G=I\alpha$

Calculation:

Friction force:

$\begin{array}{l}f={\mu }_{k}N\\ f={\mu }_{k}mg\end{array}$

Sum of horizontal forces:

$\begin{array}{l}\sum F_x=ma\\ f=ma\\ {\mu }_{k}mg=ma\\ a={\mu }_{k}g\to \end{array}$

Sum of moments about mass center:

$\begin{array}{l}\sum M_G=I\alpha \\ f\times r=m{k}^2\alpha \\ {\mu }_{k}mg\times r=m{k}^2\alpha \text{ for sphere }{k}^2=\frac{2}{5}{r}^2\\ \alpha =\frac{5{\mu }_{k}g}{2r}{\text{rad/s}}^{\text{2}}\end{array}$

Velocity equation:

$\begin{array}{l}{v}_{}=v_0-at\leftarrow \\ v_{}=v_0-{\mu }_{k}gt\leftarrow \end{array}$

Angular velocity equation:

$\begin{array}{l}\omega ={\omega }_0-\alpha t\\ \omega =\frac{5{\mu }_{k}g}{2r}t\end{array}$

Velocity of contact point:

$\begin{array}{l}{v}_C=-v+r\omega \\ v_C=-v+r(\frac{5{\mu }_{k}g}{2r}t)\\ v_C=-v+(\frac{5{\mu }_{k}g}{2}t)\end{array}$

$\begin{array}{l}{\text{when t=t}}_{\text{1}},{\text{ v=0 so v}}_{\text{c}}{\text{=v}}_{\text{1}}\\ v_1=(\frac{5{\mu }_{k}g{t}_1}{2})\\ t_1=\frac{2v_1}{5{\mu }_{k}g}\end{array}$

Conclusion:

Thus we get,

Value of time $t_1=\frac{2v_1}{5{\mu }_{k}g}$

To determine

(c)

The distance moved by sphere relative of ground.

Answer to Problem 16.74P

Distance moved $s_1=\frac{2v_1{}^2}{25{\mu }_{k}g}$

Explanation of Solution

Given information:

Mass $m$

Radius $r$

Belt velocity $v_1$

Ball velocity $v_0$

Friction coefficient ${\mu }_k$

Concept used:

Following formula is used-

1. Sum of horizontal forces, $\sum F_x=ma$

2. Sum of moments about mass center, $\sum M_G=I\alpha$

Calculation:

Friction force,

$\begin{array}{l}f={\mu }_{k}N\\ f={\mu }_{k}mg\end{array}$

Sum of horizontal forces,

$\begin{array}{l}\sum F_x=ma\\ f=ma\\ {\mu }_{k}mg=ma\\ a={\mu }_{k}g\to \end{array}$

Sum of moments about mass center,

$\begin{array}{l}\sum M_G=I\alpha \\ f\times r=m{k}^2\alpha \\ {\mu }_{k}mg\times r=m{k}^2\alpha \text{ for sphere }{k}^2=\frac{2}{5}{r}^2\\ \alpha =\frac{5{\mu }_{k}g}{2r}{\text{rad/s}}^{\text{2}}\end{array}$

Velocity equation,

$\begin{array}{l}{v}_{}=v_0-at\leftarrow \\ v_{}=v_0-{\mu }_{k}gt\leftarrow \end{array}$

Angular velocity equation,

$\begin{array}{l}\omega ={\omega }_0-\alpha t\\ \omega =\frac{5{\mu }_{k}g}{2r}t\end{array}$

Velocity of contact point,

$\begin{array}{l}{v}_C=-v+r\omega \\ v_C=-v+r(\frac{5{\mu }_{k}g}{2r}t)\\ v_C=-v+(\frac{5{\mu }_{k}g}{2}t)\end{array}$

$\begin{array}{l}{\text{when t=t}}_{\text{1}},{\text{ v=0 so v}}_{\text{c}}{\text{=v}}_{\text{1}}\\ v_1=(\frac{5{\mu }_{k}g{t}_1}{2})\\ t_1=\frac{2v_1}{5{\mu }_{k}g}\end{array}$

Distance moved,

$\begin{array}{l}{s}_1=v_0{t}_1-\frac{1}{2}{\mu }_{k}g{t}_1{}^2\\ s_1=\frac{2v_1}{5}\times \frac{2v_1}{5{\mu }_{k}g}-\frac{1}{2}\times {\mu }_{k}g\times {\left(\frac{2v_1}{5{\mu }_{k}g}\right)}^2\\ s_1=\frac{2v_1{}^2}{25{\mu }_{k}g}\end{array}$

Conclusion:

Thus we get,

Distance moved $s_1=\frac{2v_1{}^2}{25{\mu }_{k}g}.$