Question

A box contains a uniform disk of mass M and radius R that is pivoted on a low-friction axle through its centre (Figure 12.58). A block of mass m is pressed against the disk by a spring, so that the block acts like a brake, making the disk hard to turn. The box and the spring have negligible mass. A string is wrapped around the disk (out of the way of the brake) and passes through a hole in the box. A force of constant magnitude F acts on the end of the string. The motion takes place in outer space. At time ${\mathbf{t}}_{\mathbf{i}}$the speed of the box is ${\mathbf{v}}_{\mathbf{i}}$, and the rotationalspeed of the disk is ${\mathbf{\omega }}_{\mathbf{i}}$. At time ${\mathbf{t}}_{\mathbf{f}}$the box has moved a distance x, and the end of the string has moved a longer distance d, as shown.

(a) At time ${\mathbf{t}}_{\mathbf{f}}$, what is the speed ${\mathbf{v}}_{\mathbf{f}}$of the box? (b) During this process, the brake exerts a tangential friction force of magnitude f. At time ${\mathbf{t}}_{\mathbf{f}}$, what is the angular speed ${\mathbf{\omega }}_{\mathbf{f}}$of the disk? (c) At time ${\mathbf{t}}_{\mathbf{f}}$, assume that you know (from part b) the rotational speed ${\mathbf{\omega }}_{\mathbf{f}}$of the disk. From time ${\mathbf{t}}_{\mathbf{i}}$to time ${\mathbf{t}}_{\mathbf{f}}$, what is the increase in thermal energy of the apparatus? (d) Suppose that the increase in thermal energy in part (c) is $\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathbf{J}$. The disk and brake are made of iron, and their total mass is $\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{kg}$. At time ${\mathbf{t}}_{\mathbf{i}}$their temperature was . At time , what is their approximate temperature?

Short answer

1. The final velocity is $v_f=\sqrt{{v^2}_i+\frac{2x}{M}}F$.
2. The final angular speed of the disk is ${\omega }_f=\sqrt{{\omega }_i^2+\frac{2}{I}(\left(d-x\right).R.f-∆E_{thermal})}$.
3. The increase in thermal energy of the apparatus is $∆E_{thermal}=\frac{1}{2}I\left({\omega }_i^2-{\omega }_f^2\right)+\left(d-x\right).R.f.$.
4. The final temperature of disk and brake is 500.15 K

Step-by-step solution

Identification of given data

The given data can be listed below,

• The increase in thermal energy is,$Q=8\times {10}^4\mathrm{J}$ .
• Total mass of brake and disk is, $m_{d.b}=1.2\mathrm{kg}$.
• At initial time the temperature of disk and brake system is,$T_i=350\mathrm{K}$

Concept/Significance of rotational energy.

The kinetic energy associated with an item's rotation, also known as angular kinetic energy, is a component of the overall kinetic energy of the object.

The moment of inertia of an item becomes significant when looking at rotational energy independently around its axis of rotation.

(a) Determination of the speed  of the box

The work done on point particle is given by,

$W=x.F$

Here, x is the distance and F is the force applied on point particle system.

Final energy of the system is given by,

$$\begin{gathered} k_f=K_i+x.F \\ \frac{1}{2}M{v}_f^2=\frac{1}{2}M{v}_i^2+x.F \end{gathered}$$

Solve above equation for final velocity,

$$\begin{gathered} v_f^2=v_i^2+\frac{2x}{M}F \\ {v}_f=\sqrt{v_i^2+\frac{2x}{M}F} \end{gathered}$$

Thus, the final velocity is $v_f=\sqrt{v_i^2+\frac{2x}{M}F}$.

(b) Determination of the angular speed ωf  of the disk

The magnitude of the torque friction is given by,

$\tau =R.f$

Here, R is the distance between tangential friction and disk, and f is the tangential friction.

The rotational work done on the particle is given by,

$W_R=\left(d-x\right).R.f$

Here,$\left(d-x\right)$is the distance that string travels relative to disk,R is the distance between tangential friction and disk, and f is the tangential friction.

The rotational energy is given by,

$E_f=E_i+W_R+E_{thermal}$

Here, $E_i$is the initial rotational energy whose value is $\frac{1}{2}/{\omega }_i^2$, $W_R$ is the rotational work done and $E_{thermal}$is the thermal energy due to friction.

Substitute all values in the above,

$$\begin{gathered} \frac{1}{2}/{\omega }_f^2+E_{f,thermal}=\frac{1}{2}/{\omega }_f^2+\left(d-x\right).R.f+E_{i,thermal} \\ =\frac{1}{2}/{\omega }_f^2=\frac{1}{2}/{\omega }_i^2+\left(d-x\right).R.f-E_{i,thermal} \end{gathered}$$

The solution of above equation for final angular speed is given by,

localid="1657868535920" $$\begin{gathered} {\omega }_f^2=\frac{2}{1}\left[\frac{1}{2}/{\omega }_i^2+\left(d-x\right).R.f-∆E_{thermal}\right] \\ {\omega }_f=\sqrt{{\omega }_i=\frac{2}{1}\left[{\omega }_i^2+(\left(d-x\right).R.f-∆E_{thermal}\right]} \end{gathered}$$

Thus, the final angular speed of the disk is localid="1657868549031" ${\omega }_f=\sqrt{{\omega }_i=\frac{2}{1}\left[{\omega }_i^2+(\left(d-x\right).R.f-∆E_{thermal}\right]}$

(c) Determination of the increase in thermal energy of the apparatus

The rotational energy of the disk is given by,

$E_f=E_i+W_R+E_{thermal}$

Here, is the initial rotational energy whose value is $\frac{1}{2}/{\omega }_i^2$, is the rotational work done and$E_{thermal}$is the thermal energy due to friction.

Substitute values in the above,

role="math" localid="1657869838373" $\frac{1}{2}/{\omega }_i^2+E_{thermal}=\frac{1}{2}/{\omega }_i^2+\left(d-x\right).R.f+E_{i,thermal}$

The increase on thermal energy is given by,

role="math" localid="1657870147400" $$\begin{gathered} E_{f,thermal}-E_{i,thermal}=\frac{1}{2}/{\omega }_i^2-\frac{1}{2}/{\omega }_f^2+\left(d-x\right).R.f \\ ∆E_{i,thermal}=\frac{1}{2}/\left({\omega }_i^2-{\omega }_f^2\right)+\left(d-x\right).R.f \end{gathered}$$

Thus, the increase in thermal energy of the apparatus is $∆E_{i,thermal}=\frac{1}{2}/\left({\omega }_i^2-{\omega }_f^2\right)+\left(d-x\right).R.f$ .

(d) Determination of the approximate temperature of disk and brakes

The heat which is equal to change in thermal energy is given by,

$Q=mc\left(T_f-T_i\right)$

Here, m is the total mass of the system, c is the specific heat of iron whose value is 444 J/kg , $T_f$is the final temperature of disk and brake system and is the initial temperature of the system.

The final temperature of the disk and brake is given by,

$T_f=T_i+\frac{Q}{m_{d,b}C}$

Substitute all the values in the above,

$$\begin{gathered} T_f=\left(350\mathrm{K}\right)+\frac{8\times {10}^4\mathrm{J}}{m_{d,b}C} \\ =350+150.15 \\ =500.15\mathrm{K} \end{gathered}$$

Thus, the final temperature of disk and brake is 500.15 K