Chapter 10, Problem 10.55P

Review. An object with a mass of m = 5.10 kg is attached to the free end of a light string wrapped around a reel of radius R = 0.250 m and mass M = 3.00 kg. The reel is a solid disk, free to rotate in a vertical plane about the horizontal axis passing through its center as shown in Figure P10.29. The suspended object is released from rest 6.00 m above the floor. Determine (a) the tension in the string, (b) the acceleration of the object, and (c) the speed with which the object hits the floor. (d) Verify your answer to part (c) by using the isolated system (energy) model.

Figure P10.29

To determine

The tension in the string.

Answer to Problem 10.55P

The tension in the string is $11.4\text{N}$.

Explanation of Solution

The mass of the object is $5.10\text{kg}$, the mass of the solid disk reel is $3.00\text{kg}$ and the radius of the disk is $0.250\text{m}$. The final height of the object from the floor is $6.0\text{m}$.

Formula to calculate the torque produce by the tension in the string is,

    $\tau =TR$        (I)

Here, $\tau$ is the torque produced by the tension, $T$ is the tension in the string and $R$ is the radius of the solid disk.

Formula to calculate the torque due to the rotation of the disk is,

    $\tau =I\alpha$        (II)

Since the torque produce by the tension is cause the disk to rotate hence both the torque must be equal.

Equate equation (I) and equation (II).

    $TR=I\alpha$        (III)

Formula to calculate the moment of inertia of the solid disk is,

    $I=\frac{M{R}^2}{2}$

Here, $I$ is the moment of inertia of the solid disk, $M$ is the mass of the solid disk and $R$ is the radius of the solid disk.

Formula to calculate the angular acceleration of the solid disk is,

    $\alpha =\frac{a}{R}$

Here, $\alpha$ is the angular acceleration of the disk and $a$ is the linear acceleration of the object.

Substitute $\frac{M{R}^2}{2}$ for $I$ and $\frac{a}{R}$ for $\alpha$ in equation (III)

    $\begin{array}{c}TR=\frac{M{R}^2}{2}\times \frac{a}{R}\\ T=\frac{Ma}{2}\\ a=\frac{2T}{M}\end{array}$        (IV)

Formula to calculate the net vertical force act on the block is,

    $mg-T=ma$

Here, $m$ is the mass of the object and $g$ is the acceleration due to gravity.

Substitute $\frac{2T}{M}$ for $a$ in above equation.

    $mg-T=m\left(\frac{2T}{M}\right)$

Rearrange the equation for $T$,

    $\begin{array}{c}mg=\left(\frac{2m}{M}+1\right)T\\ T=\frac{mg}{\frac{2m}{M}+1}\end{array}$

Substitute $5.10\text{kg}$ for $m$, $3.00\text{kg}$ for $M$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ to find $T$.

    $\begin{array}{c}T=\frac{5.10\text{kg}\times 9.8\text{m}/{\text{s}}^{\text{2}}}{\frac{2\times 5.10\text{kg}}{3.00\text{kg}}+1}\\ =\frac{49.98\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}{4.4}\times \frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\\ =11.4\text{N}\end{array}$

Conclusion:

Therefore, the tension in the string is $11.4\text{N}$.

To determine

The acceleration of the object.

Answer to Problem 10.55P

The acceleration of the object is $7.6\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

The mass of the object is $5.10\text{kg}$, the mass of the solid disk reel is $3.00\text{kg}$ and the radius of the disk is $0.250\text{m}$. The final height of the object from the floor is $6.0\text{m}$.

Formula to calculate the acceleration of the object from equation (IV) is,

    $a=\frac{2T}{M}$

Substitute for $11.4\text{N}$ and $3.00\text{kg}$ for $M$ in above equation to find $a$.

    $\begin{array}{c}a=\frac{2\times 11.4\text{N}}{3.00\text{kg}}\times \frac{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}{1\text{N}}\\ =7.6\text{m}/{\text{s}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the acceleration of the object $7.6\text{m}/{\text{s}}^{\text{2}}$.

To determine

The speed with which the object hits the floor.

Answer to Problem 10.55P

The speed with which the object hits the floor is $9.5\text{m}/\text{s}$.

Explanation of Solution

The mass of the object is $5.10\text{kg}$, the mass of the solid disk reel is $3.00\text{kg}$ and the radius of the disk is $0.250\text{m}$. The final height of the object from the floor is $6.0\text{m}$.

Formula to calculate the speed of the object using Newton’s equation is,

    $v_{\text{f}}^2=v_{\text{i}}^2+2ah$

Here, $v_{\text{f}}$ is the final speed of the object, $v_{\text{i}}$ is the initial speed of the object and $h$ is the vertical distance travel by the object

Substitute $0$ for $v_{\text{i}}$, $7.6\text{m}/{\text{s}}^{\text{2}}$ for $a$ and $6.0\text{m}$ for $h$ in above equation to find $v_{\text{f}}$.

    $\begin{array}{c}{v}_{\text{f}}^2=0+2\times 7.6\text{m}/{\text{s}}^{\text{2}}\times 6.0\text{m}\\ v_{\text{f}}=\sqrt{91.2{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\\ =9.5\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the speed with which the object hits the floor is $9.5\text{m}/\text{s}$.

To determine

The answer in part (a) by use of isolated system (energy) model.

Answer to Problem 10.55P

The velocity of the object in part (c) and in part (d) are comes out same. Hence the answer to part (c) is verified by using the isolated system model.

Explanation of Solution

The mass of the object is $5.10\text{kg}$, the mass of the solid disk reel is $3.00\text{kg}$ and the radius of the disk is $0.250\text{m}$. The final height of the object from the floor is $6.0\text{m}$.

The initial rotational and translational kinetic energy of the object is zero since the object is initially at rest. Also the final potential energy is zero.

The law of conservation of energy for the system is given as,

    $U_{\text{i}}=K_{\text{f}}+K_{\text{r}}$        (V)

Here, $U_{\text{i}}$ is the initial potential energy of the object, $K_{\text{f}}$ is the final kinetic energy of the object and $K_{\text{r}}$ is the rotational kinetic energy of the sold disk reel.

Formula to calculate the initial potential energy of the object is,

    $U_{\text{i}}=mgh$

Formula to calculate the final kinetic energy of the object is,

    $K_{\text{f}}=\frac{1}{2}m{v}_{\text{f}}^2$

Formula to calculate the rotational kinetic energy of the sold disk reel is,

    $K_{\text{r}}=\frac{1}{2}I{\left(\frac{v_{\text{f}}}{R}\right)}^2$

Substitute $\frac{M{R}^2}{2}$ for $I$     in above equation.

    $\begin{array}{c}{K}_{\text{r}}=\frac{1}{2}\left(\frac{M{R}^2}{2}\right){\left(\frac{v_{\text{f}}}{R}\right)}^2\\ =\frac{1}{4}M{v}_{\text{f}}^2\end{array}$

Substitute $mgh$ for $U_{\text{i}}$, $\frac{1}{2}m{v}_{\text{f}}^2$ for $K_{\text{f}}$ and $\frac{1}{4}M{v}_{\text{f}}^2$ for $K_{\text{r}}$ in equation (V).

    $mgh=\frac{1}{2}m{v}_{\text{f}}^2+\frac{1}{4}M{v}_{\text{f}}^2$

Rearrange the equation for $v_{\text{f}}$.

    $\begin{array}{c}mgh=\frac{1}{2}\left(m+\frac{M}{2}\right)v_{\text{f}}^2\\ v_{\text{f}}=\sqrt{\frac{4mgh}{2m+M}}\end{array}$

Conclusion:

Substitute $5.10\text{kg}$ for $m$, $3.00\text{kg}$ for $M$, $6.0\text{m}$ for $h$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ to find $v_{\text{f}}$.

    $\begin{array}{c}{v}_{\text{f}}=\sqrt{\frac{4\times 5.10\text{kg}\times 9.8\text{m}/{\text{s}}^{\text{2}}\times 6.0\text{m}}{2\times 5.10\text{kg}+3.0\text{kg}}}\\ =\sqrt{91{{\text{m}}^2/\text{s}}^2}\\ =9.54\text{m}/\text{s}\end{array}$

Therefore, the velocity of the object in part (c) and in part (d) are comes out same. Hence the answer to part (c) is verified by using the isolated system model.