Chapter 10, Problem 45P

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 P10.45. The suspended object is released from rest 6.00 m above the floor. Determine

1. (a) the tension in the string,
2. (b) the acceleration of the object, and
3. (c) the speed with which the object hits the floor.
4. (d) Verify your answer to part (c) by using the isolated system (energy) model.

Figure P10.45

To determine

The tension acting on the string.

Answer to Problem 45P

The tension acting on the string is $\underset{\_}{11.4\text{N}}$.

Explanation of Solution

Consider the figure given below.

The forces acting on the reel are given in the figure 1. The gravitational force acting on the reel is $mg,$ $n$ is the normal force exerted by the axle, and tension is the only force responsible for the torque causing rotation of the reel.

Write the expression for the gravitational force.

  $F_g=mg$        (I)

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

The reel can be considered as a uniform disk, let $I$ be the moment of inertia of the reel.

Write the expression for the moment of inertia of the reel.

  $I=\frac{1}{2}M{R}^2$        (II)

Here, $M$ is the mass of the reel, $R$ is the radius of the reel.

Write the expression for torque about the axis of rotation.

  ${\displaystyle \sum \tau }=n{r}_1+F_{gp}{r}_2+T{r}_3$        (III)

Here, $r_1$ is the distance from axis of rotation to the normal force, $r_2$ is the distance between the axis of rotation and force $F_{gp}$, $T$ is the tension, $r_3$ is the distance between axis of rotation and $T$.

Equate $I\alpha$ for $T$ in equation (III).

  $I\alpha =n{r}_1+F_{gp}{r}_2+T{r}_3$        (IV)

Here, $\alpha$ is the angular acceleration.

Substitute, $\frac{a_t}{r}$ for $\alpha$ in equation (IV).

  $I\left(\frac{a_t}{r}\right)=n{r}_1+F_{gp}{r}_2+T{r}_3$        (V)

Write the expression of force in downward direction.

  $\begin{array}{c}{\displaystyle \sum F_y}=mg-T\\ =ma\end{array}$        (VI)

Conclusion:

Substitute, $5.10\text{kg}$ for $m$, $9.80\text{m}/{\text{s}}^2$ for $g$ in equation (I).

  $\begin{array}{c}{F}_g=5.10\text{kg}\times 9.80\text{m}/{\text{s}}^2\\ =50.0\text{N}\end{array}$

Substitute, $3.00\text{kg}$ for $M$, $0.250\text{m}$ for $R$ in equation (II).

  $\begin{array}{c}I=\frac{1}{2}\left(3.00\text{kg}\right){\left(0.250\text{m}\right)}^2\\ =0.093\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute, $0$ for $r_1$, $0$ for $r_2$, $0.250\text{m}$ for $r_3$, $0.093\text{kg}\cdot {\text{m}}^2$ for $I$ in equation (V).

  $\begin{array}{c}0.093\text{kg}\cdot {\text{m}}^2\left(\frac{a_t}{0.250\text{m}}\right)=n\left(0\right)+F_{gp}\left(0\right)+T\left(0.250\text{m}\right)\\ =T\left(0.250\text{m}\right)\end{array}$        (VII)

Substitute, $50.0\text{N}$ for $mg$, $5.10\text{kg}$ for $m$ in equation (VI).

  $\begin{array}{c}50.0\text{N}-T=\left(5.10\text{kg}\right)a\\ T=50.0\text{N}-\left(5.10\text{kg}\right)a\end{array}$        (VIII)

Substitute, equation (VII) in equation (VIII) to obtain the value of acceleration.

  $\begin{array}{c}0.093\text{kg}\cdot {\text{m}}^2\left(\frac{a_t}{0.250\text{m}}\right)=\left[50.0\text{N}-\left(5.10\text{kg}\right)a\right]\left(0.250\text{m}\right)\\ 12.5\text{N}\cdot \text{m}-\left(1.28\text{kg}\cdot \text{m}\right)a=\left(0.375\text{kg}\cdot \text{m}\right)a\\ a=7.57\text{m}/{\text{s}}^2\end{array}$

Substitute, $7.57\text{m}/{\text{s}}^2$ for $a$ in equation (VIII).

  $\begin{array}{c}T=50.0\text{N}-\left(5.10\text{kg}\right)\left(7.57\text{m}/{\text{s}}^2\right)\\ =11.4\text{N}\end{array}$

Therefore, the tension acting on the string is $\underset{\_}{11.4\text{N}}$.

To determine

The acceleration of the object.

Answer to Problem 45P

The acceleration of the object is ${\underset{\_}{7.57\text{m}/\text{s}}}^2$.

Explanation of Solution

The acceleration is already calculated in part (a).

Conclusion:

Therefore, the acceleration of the object is $\underset{\_}{7.57{\text{m}/\text{s}}^2}$.

To determine

The speed with which object hits the ground.

Answer to Problem 45P

The speed with which object hits the ground is $\underset{\_}{9.53\text{m}/\text{s}}$, downwards.

Explanation of Solution

Write the Newton’s equation of motion for the final velocity.

  ${\upsilon }_f^2={\upsilon }_i^2+2a\left(y_f-y_i\right)$        (IX)

Here, ${\upsilon }_f$ is the final velocity, ${\upsilon }_i$ is the initial velocity, $y_f$ is the final position, $y_i$ is the initial position.

Conclusion:

Substitute, $0$ for ${\upsilon }_i$, $7.57{\text{m}/\text{s}}^2$ for $a$, and $6.00\text{m}$ for $y_f-y_i$ in equation (IX).

  $\begin{array}{c}{\upsilon }_f^2=0+2\left(7.57{\text{m}/\text{s}}^2\right)\left(6.00\text{m}\right)\\ {\upsilon }_f=9.53\text{m}/\text{s}\end{array}$

Therefore, the speed with which object hits the ground is $\underset{\_}{9.53\text{m}/\text{s}}$, downwards.

To determine

To verify the value of speed of the object using isolated system model.

Answer to Problem 45P

The value of acceleration obtained using both methods are same.

Explanation of Solution

Write the equation for the conservation energy in case of the given system.

  ${\left(K_1+K_2+U_g\right)}_i={\left(K_1+K_2+U_g\right)}_f$        (X)

Here, $K_1$ is the kinetic energy of the falling object, $K_2$ is the kinetic energy of the reel, $U_g$ is the gravitational potential energy of the falling object.

Substitute, $0$ for $K_{1i}$, and $K_{2i}$, $m_{1}g{y}_{1i}$ for $U_{gi}$, $0$ for $U_{gf}$, $\frac{1}{2}{m}_1{\upsilon }_{1f}^2$ for $K_{1f}$, and $\frac{1}{2}{I}_2{\omega }_{2f}^2$ for $K_{2f}$ in equation (X) and rearrange to obtain an expression for $\upsilon$.

  $\begin{array}{c}\left(0+0+m_{1}g{y}_{1i}\right)=\left(\frac{1}{2}{m}_1{\upsilon }_{1f}^2+\frac{1}{2}{I}_2{\omega }_{2f}^2+0\right)\\ mg{y}_i=\frac{1}{2}m{\upsilon }^2+\frac{1}{2}I{\omega }^2\end{array}$        (XI)

Substitute, $\frac{\upsilon }{R}$ for $\omega$ in equation (XI).

  $\begin{array}{r}mg{y}_i=\frac{1}{2}m{\upsilon }^2+\frac{1}{2}I{\left(\frac{\upsilon }{R}\right)}^2\\ \upsilon =\sqrt{\frac{2mg{y}_i}{m+\left(I/R^2\right)}}\end{array}$        (XII)

Conclusion:

Substitute, $5.10\text{kg}$ for $m$, $9.80\text{m}/{\text{s}}^2$ for $g$, $6.00\text{m}$ for $y_i$, $0.250\text{m}$ for $R$, $0.093\text{kg}\cdot {\text{m}}^2$ for $I$ in equation (XII).

  $\begin{array}{c}\upsilon =\sqrt{\frac{2\left(5.10\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)\left(6.00\text{m}\right)}{5.10\text{kg}+\left(0.093\text{kg}\cdot {\text{m}}^2/0.250{\text{m}}^2\right)}}\\ =9.53\text{m}/\text{s}\end{array}$

Therefore, the value of acceleration obtained using both methods are same.