Chapter 5, Problem 5.76AP

A 1.00-kg glider on a horizontal air track is pulled by a string at an angle θ. The taut string runs over a pulley and is attached to a hanging object of mass 0.500 kg as shown in Figure P5.40. (a) Show that the speed v_x of the glider and the speed v_y of the hanging object are related by v_x = uv_y, where u = z(z² − h₀²)^−1/2. (b) The glider is released from rest. Show that at that instant the acceleration a_x of the glider and the acceleration a_y of the hanging object are related by a_x = ua_y. (c) Find the tension in the string at the instant the glider is released for h₀ = 80.0 cm and θ = 30.0°.

Figure P5.40

To determine

The relation between the speed of the glider and the speed of the hanging object.

Answer to Problem 5.76AP

The relation between the speed of the glider and the speed of the hanging object is $v_{\text{x}}=u{v}_y$ where $u=z{\left(z^2-h_0^2\right)}^{-\left(\frac{1}{2}\right)}$.

Explanation of Solution

The mass of the glider is $1.00\text{kg}$, the angle between the string and horizontal is $\theta$, the mass of the hanging object is $0.500\text{kg}$.

The free body diagram of the given case is as shown below.

Figure (1)

Form the above figure (1).

Write the expression for the length of the string using Pythagorean Theorem,

    $z^2=x^2+{\left(h_0\right)}^2$

Here, $z$ is the length of string, $x$ is the distance of the glider on the ruler scale and $h_0$ is the string length that is holding the hanging object.

Rearrange the above equation for $x$.

    $x={\left(z^2-{\left(h_0\right)}^2\right)}^{\frac{1}{2}}$

Write the expression for the speed of the glider

    $v_{\text{x}}=\frac{dx}{dt}$

Here, $v_{\text{x}}$ is the speed of the glider.

Substitute ${\left(z^2-{\left(h_0\right)}^2\right)}^{\frac{1}{2}}$ for $x$ in the above equation.

    $\begin{array}{c}{v}_{\text{x}}=\frac{d}{dt}\left({\left(z^2-{\left(h_0\right)}^2\right)}^{\frac{1}{2}}\right)\\ =\frac{1}{2}{\left(z^2-{\left(h_0\right)}^2\right)}^{-\left(\frac{1}{2}\right)}2z\frac{dz}{dt}\end{array}$                                                                       (I)

The term $\frac{dz}{dt}$ in the above expression is the rate of the string passing over the pulley.

Write the expression for the speed of the hanging object.

    $v_{\text{y}}=\frac{dz}{dt}$

Here, $v_{\text{y}}$ is the speed of the hanging object.

Substitute $v_{\text{y}}$ for $\frac{dz}{dt}$ in the equation (1).

    $\begin{array}{c}{v}_{\text{x}}=\frac{1}{2}{\left(z^2-{\left(h_0\right)}^2\right)}^{-\left(\frac{1}{2}\right)}2z\left(v_{\text{y}}\right)\\ =z{\left(z^2-{\left(h_0\right)}^2\right)}^{-\left(\frac{1}{2}\right)}\left(v_{\text{y}}\right)\end{array}$

Substitute $u$ for $z{\left(z^2-{\left(h_0\right)}^2\right)}^{-\left(\frac{1}{2}\right)}$ in the above equation.

    $v_{\text{x}}=u\left(v_{\text{y}}\right)$                                                                                                  (II)

Conclusion:

Therefore, the relation between the speed of the glider and the speed of the hanging object is $v_{\text{x}}=u{v}_y$ where $u=z{\left(z^2-h_0^2\right)}^{-\left(\frac{1}{2}\right)}$.

To determine

The relation between the acceleration of the glider and the speed of the hanging object.

Answer to Problem 5.76AP

The relation between the acceleration of the glider and the speed of the hanging object is $a_{\text{x}}=u{a}_y$.

Explanation of Solution

From equation (2), the relation of $v_{\text{x}}$ and $v_{\text{y}}$ is given as,

    $v_{\text{x}}=u\left(v_{\text{y}}\right)$

Write the expression for the acceleration of the glider

    $a_{\text{x}}=\frac{d}{dt}{v}_{\text{x}}$

Substitute $u\left(v_{\text{y}}\right)$ for $v_{\text{x}}$ in the above equation.

    $\begin{array}{c}{a}_{\text{x}}=\frac{d}{dt}\left[u\left(v_{\text{y}}\right)\right]\\ =u\frac{d}{dt}\left(v_{\text{y}}\right)+v_{\text{y}}\frac{du}{dt}\end{array}$

The initial velocity of the hanging object is zero.

Substitute $0$ for $v_{\text{y}}$ and $a_{\text{y}}$ for $\frac{d}{dt}\left(v_{\text{y}}\right)$ in the above equation.

    $a_{\text{x}}=u{a}_{\text{y}}$

Here, $a_{\text{y}}$ is the acceleration of the hanging object.

Conclusion:

Therefore, the relation between the acceleration of the glider and the speed of the hanging object is $a_{\text{x}}=u{a}_{\text{y}}$.

To determine

The tension of the string.

Answer to Problem 5.76AP

The tension of the string is $3.56\text{N}$.

Explanation of Solution

From the free body diagram in figure (1) the net direction in $x$ direction

    $z=\frac{h_0}{\sin \theta }$

From part (a) the value of $u$

    $u=z{\left(z^2-h_0^2\right)}^{-\left(\frac{1}{2}\right)}$

Substitute $\frac{h_0}{\sin \theta }$ for $z$ in the above equation.

    $u=\frac{h_0}{\sin \theta }{\left({\left(\frac{h_0}{\sin \theta }\right)}^2-h_0^2\right)}^{-\left(\frac{1}{2}\right)}$

Substitute $30.0°$ for $\theta$ and $80.0\text{cm}$ for $h_0$ in the above equation.

    $\begin{array}{c}u=\frac{80.0\text{cm}}{\sin \left(30.0°\right)}{\left({\left(\frac{80.0\text{cm}}{\sin \left(30.0°\right)}\right)}^2-{\left(80.0\text{cm}\right)}^2\right)}^{-\left(\frac{1}{2}\right)}\\ =115\text{cm}\times \frac{\text{1}\text{m}}{100\text{cm}}\\ =1.15\text{m}\end{array}$

Thus, the value of $u$  is $1.15\text{m}$.

The net force in $y$ direction

    $T-mg=-m{a}_{\text{y}}$

Here, $T$  is the tension in the wire and $m$ is the mass of the hanging block.

Substitute $0.5\text{kg}$ for $m$ and $9.8{\text{m}/\text{s}}^{\text{2}}$ for $g$.

    $T-\left(0.5\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)=-\left(0.5\text{ kg}\right)a_{\text{y}}$

Rearrange the above equation for $a_{\text{y}}$.

    $a_{\text{y}}=\left(-\frac{T}{0.5\text{kg}}\right)+9.8{\text{m}/\text{s}}^{\text{2}}$                                                                        (III)

The net force in the $x$ direction

    $T\cos \theta =M{a}_{\text{x}}$

Here, $M$ is the mass of the glider.

Form part (b) substitute $u{a}_{\text{y}}$ for $a_{\text{x}}$ in the above equation.

    $\begin{array}{c}T\cos \theta =Mu{a}_{\text{y}}\\ T=\frac{Mu{a}_{\text{y}}}{\cos \theta }\end{array}$

From equation (3) substitute $\left(-\frac{T}{0.5\text{kg}}\right)+9.8{\text{m}/\text{s}}^{\text{2}}$ for $a_{\text{y}}$ in the above equation,

    $T=\frac{Mu}{\cos \theta }\left(\left(-\frac{T}{0.5\text{kg}}\right)+9.8{\text{m}/\text{s}}^{\text{2}}\right)$

Conclusion:

Substitute $1.00\text{kg}$ for $M$, $0.500\text{kg}$ for $m$ and $1.15\text{m}$ for $u$ in the above equation.

    $\begin{array}{c}T=\frac{\left(1.00\text{kg}\right)\left(1.15\text{m}\right)}{\cos \left(30.0°\right)}\left(\left(-\frac{T}{0.5\text{kg}}\right)+9.8{\text{m}/\text{s}}^{\text{2}}\right)\\ 3.655T=13.013\\ T=3.56\text{N}\end{array}$

Therefore, the tension in the string is $3.56\text{N}$.