Chapter 5, Problem 40AP

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

To show: The speed of the glider and the speed of the hanging object are related by $v_{\text{x}}=u{v}_y$ where $u=z{\left(z^2-h_0^2\right)}^{-\left(\frac{1}{2}\right)}$.

Answer to Problem 40AP

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

Explanation of Solution

Given info: 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).

From the 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.

$h_0$ is the string length that is holding the hanging object.

Rearrange the above equation for $x$.

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

The speed of the glider is.

$v_{\text{x}}=\frac{dx}{dt}$

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

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

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

$v_{\text{y}}=\frac{dz}{dt}$

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.

$v_{\text{x}}=u\left(v_{\text{y}}\right)$ (2)

Conclusion:

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

To determine

To show: The acceleration of the glider and the speed of the hanging object are related by. $a_{\text{x}}=u{a}_y$.

Answer to Problem 40AP

The acceleration of the glider and the speed of the hanging object are related by. $a_{\text{x}}=u{a}_y$.

Explanation of Solution

Given info: 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 $0.500\text{kg}$.

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

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

The acceleration of the glider is,

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

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 is zero.

Substitute $0$ for $u$ 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 acceleration of the glider and the speed of the hanging object are related by. $a_{\text{x}}=u{a}_{\text{y}}$.

To determine

The tension of the string.

Answer to Problem 40AP

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

Explanation of Solution

Given info: The mass of the glider is $1.00\text{kg}$, the angle between the string and horizontal is $30.0°$, the mass of the hanging object is $0.500\text{kg}$.

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

$z=\frac{h_0}{\sin \theta }$

From part (a) the value of $u$ is,

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

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 is,

$T-mg=-m{a}_{\text{y}}$

Here,

$T$ is the tension in the wire.

$m$ 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 $a_{\text{y}}$.

$a_{\text{y}}=\left(-\frac{T}{0.5\text{kg}}\right)+9.8{\text{m}/\text{s}}^{\text{2}}$ (3)

The net force in the $x$ direction is,

$T\cos \theta =M{a}_{\text{x}}$

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,

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

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

$\begin{array}{c}T=\frac{\left(1.00\text{kg}\right)\left(1.15\text{m}\right)}{\cos \left(30.0°\right)}\left(\left(-\frac{T}{0.5\text{kg}}\right)+9.8{\text{m}/\text{s}}^{\text{2}}\right)\\ 3.18T=11.3\text{N}\\ T=3.56\text{N}\end{array}$

Conclusion:

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