Chapter 4, Problem 72P

(a)

To determine

The distance $D$ from the bottom of the bar where mug hit the floor.

Answer to Problem 72P

The distance $D$ from the bottom of the bar where mug hit the floor is $\underset{\_}{0.84\text{m}}$.

Explanation of Solution

Write the mathematical expression for Newton’s second law.

  ${\displaystyle \sum \dot{F}}=m\dot{a}$        (I)

Here, $m$ is the mass, $\dot{a}$ is the acceleration.

Write the kinematic equation for velocity in case of mug.

  $v_x^2=v_{0x}^2+2a_x\left(x-x_0\right)$        (II)

Here, $v_{0x}$ is the initial velocity, $v_x$ is the final velocity, $a_x$ is the acceleration, $x-x_0$ is the displacement.

After the mug leaves the bar the projectile motion can be considered.

Consider the free body diagram.

Apply Newton’s second law to the free body diagram.

Write the expression for the $x$ component of force.

  $\begin{array}{c}{\displaystyle \sum F_x}=-F_{friction}\\ =ma\end{array}$        (III)

Substitute, ${\mu }_{k}g$ for $F_{friction}$ in equation (III), and rearrange to obtain an expression for mass.

  $\begin{array}{c}{F}_{friction}={\mu }_{k}N\\ =ma\\ m=\frac{{\mu }_{k}N}{a}\end{array}$        (IV)

Write the expression for $y$ component of force.

  $\begin{array}{c}{\displaystyle \sum F_y}=N-mg\\ =0\end{array}$        (IV)

Here, $N$ is the normal force, $g$ is the acceleration due to gravity.

Substitute equation (III) in(IV).

  $\begin{array}{c}N-mg=0\\ N=\frac{{\mu }_{k}N}{a}g\\ a={\mu }_{k}g\end{array}$        (V)

Here, ${\mu }_k$ is the coefficient of kinetic friction.

Write the kinematic equation for displacement in $y$ direction.

  $y=y_0+v_{0y}t-\frac{1}{2}g{t}^2$        (VI)

Substitute, $0$ for $v_{0y}$, and $y_0$ in equation (VI), and rearrange to obtain an expression for $t$.

  $\begin{array}{c}y=0+\left(0\right)t-\frac{1}{2}g{t}^2\\ =-\frac{1}{2}g{t}^2\\ t=\sqrt{\frac{2\left(-y\right)}{g}}\end{array}$        (VII)

Write the kinematic equation for displacement in $x$ direction.

  $x=x_0+v_{0x}t$        (VIII)

Conclusion:

Substitute, $0.08$ for ${\mu }_k$, and $-9.8\text{m}/{\text{s}}^2$ for $g$ in equation (V).

  $\begin{array}{c}a=\left(0.08\right)\left(-9.8\text{m}/{\text{s}}^2\right)\\ =-0.784\text{m}/{\text{s}}^2\end{array}$

Substitute, $2.5\text{m}/\text{s}$ for $v_{0x}$, $0.784\text{m}/{\text{s}}^2$ for $a_x$, $2.0\text{m}$ for $x-x_0$ in equation (II).

  $\begin{array}{c}{v}_x^2={\left(2.5\text{m}/\text{s}\right)}^2+2\left(0.784\text{m}/{\text{s}}^2\right)\left(2.0\text{m}\right)\\ ={\left(3.11\text{m}/\text{s}\right)}^2\\ v_x=1.76\text{m}/\text{s}\end{array}$

Substitute, $1.1\text{m}$ for $y$, and $-9.8\text{m}/{\text{s}}^2$ for $g$ in equation (VIII).

  $\begin{array}{c}t=\sqrt{\frac{2\left(-1.1\text{m}\right)}{-9.8\text{m}/{\text{s}}^2}}\\ =0.474\text{s}\end{array}$

Substitute, $1.76\text{m}/\text{s}$ for $v_{0x}$, $0$ for $x_0$, and $0.474\text{s}$ for $t$ in equation (VIII).

  $\begin{array}{c}x=0+\left(1.76\text{m}/\text{s}\right)\left(0.474\text{s}\right)\\ =0.84\text{m}\end{array}$

Therefore, the distance $D$ from the bottom of the bar where mug hit the floor is $\underset{\_}{0.84\text{m}}$.

(b)

To determine

The velocity of the mug.

Answer to Problem 72P

The velocity of the mug is $\underset{\_}{50\text{m}/\text{s}}$, at angle $69°$ below the $x$ axis.

Explanation of Solution

Write the expression for velocity along $y$ direction

  $v_y=v_{0y}-gt$        (IX)

Here the $x$ component of velocity is $1.76\text{m}/\text{s}$.

Write the expression for the magnitude of the velocity.’

  $v=\sqrt{v_x^2+v_y^2}$        (X)

Write the expression for the direction of velocity.

  $\theta ={\tan }^{-1}\left(\frac{v_y}{v_x}\right)$        (XI)

Conclusion:

Substitute, $0$ for $v_0$, $9.8\text{m}/{\text{s}}^2$ for $g$, and $0.474\text{s}$ $t$ in equation (IX).

  $\begin{array}{c}{v}_y=0-\left(9.8\text{m}/{\text{s}}^2\right)\left(0.474\text{s}\right)\\ =-4.65\text{m}/\text{s}\end{array}$

Substitute, $1.76\text{m}/\text{s}$ for $v_x$, and $-4.65\text{m}/\text{s}$ for $v_y$ in equation (X) to find the magnitude of the velocity.

  $\begin{array}{c}v=\sqrt{\left(1.76\text{m}/\text{s}\right)+\left(-4.65\text{m}/\text{s}\right)}\\ =5.0\text{m}/\text{s}\end{array}$

Substitute, $1.76\text{m}/\text{s}$ for $v_x$, and $-4.65\text{m}/\text{s}$ for $v_y$ in equation (XI) to find the direction of the velocity.

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-4.65\text{m}/\text{s}}{1.76\text{m}/\text{s}}\right)\\ =-69°\end{array}$

Therefore, the velocity of the mug is $\underset{\_}{50\text{m}/\text{s}}$, at angle $69°$ below the $x$ axis.

(c)

To determine

The velocity time graph for both $x$ and $y$ directions.

Answer to Problem 72P

The velocity time graph for both $x$ and $y$ directions is given below.

Explanation of Solution

Write the expression for the $x$ component of velocity.

  $v_x=v_{0x}+a{x}_{}t$        (XII)

Conclusion:

Substitute, $1.77\text{m}/\text{s}$ for $v_{0x}$, $2.5\text{m}/\text{s}$ for $v_x$, $-0.78\text{m}/{\text{s}}^2$ for $a_x$, in equation (XII) to find the time.

  $\begin{array}{c}2.5\text{m}/\text{s}=1.77\text{m}/\text{s}+\left(-0.78\text{m}/{\text{s}}^2\right)t\\ t=\frac{2.5\text{m}/\text{s}-1.77\text{m}/\text{s}}{-0.78\text{m}/{\text{s}}^2}\\ =0.94\text{s}\end{array}$

The velocity time graph for velocity in $x$ direction.

The velocity time graph for velocity in $y$ direction.

Therefore, the velocity of the mug is $\underset{\_}{50\text{m}/\text{s}}$, at angle $69°$ below the $x$ axis.