Chapter 8, Problem 83GP

(a)

To determine

To Find: The angular momentum of $M_A$ .

Answer to Problem 83GP

The value of angular momentum is $7.78\text{kg}\cdot {\text{m}}^{\text{2}}\text{/s}$ .

Explanation of Solution

Given:

Mass of first cylindrical plate, $M_A=6\mathrm{kg}$

Mass of second cylindrical plate, $M_B=9\mathrm{kg}$

The angular velocity of the first plate, ${\omega }_1=7.2\text{rad/s}$

The radius of each clutch, $R=0.60\text{m}$

Time taken, $\Delta t=2\mathrm{s}$

Formula used:

Angular momentum $L_A=I_A{\omega }_A=\frac{1}{2}{M}_A{R}^2{\omega }_A\mathrm{....}(1)$

Where, $I_A$ is moment of inertia.

  ${\omega }_A$ is angular velocity.

  $M_A$ is mass of disc.

  $R$ is radius of disc.

Calculation:

Substitute the given values in equation (1):

  $\begin{array}{l}{L}_A=\frac{1}{2}(6){(}^{0.60}(7.2)\\ =7.78\mathrm{kg}\cdot m^2/s\end{array}$

Conclusion:

The value of angular momentum is $7.78\text{kg}\cdot {\text{m}}^{\text{2}}\text{/s}$ .

(b)

To determine

To Find: The torque required to accelerate the first plate.

Answer to Problem 83GP

The torque required to accelerate the first plate is $3.9\mathrm{N}\cdot m$

Explanation of Solution

Given:

The angular momentum of the first plate, $L_A=7.78\text{kg}\cdot {\text{m}}^{\text{2}}\text{/s}$

Time is taken, $\Delta t=2\mathrm{s}$

Formula used:

Let the torque required to accelerate the first pate is $\tau$

Torque is the rate of change of angular momentum $\tau =\frac{\Delta L}{\Delta t}.$

Change of angular momentum $\Delta L=L_f-L_i.$

Calculation:

  $\begin{array}{l}\tau =\frac{7.78-0}{2}\\ =\frac{7.78}{2}\\ =3.89\mathrm{N}.m\end{array}$

Conclusion:

The value of torque is $3.89\mathrm{N}.m$ .

(c)

To determine

To Find: The angular velocity of second plate.

Answer to Problem 83GP

The angular velocity of second plate is $2.9\text{rad/s}\text{.}$

Explanation of Solution

Given:

Mass of first cylindrical plate, $M_A=6\mathrm{kg}$

Mass of second cylindrical plate, $M_B=9\mathrm{kg}$

The angular velocity of the first plate, ${\omega }_1=7.2\text{rad/s}$

The radius of each clutch, $R=0.60\mathrm{m}$

Formula used:

When there is no external torque then the conservation of angular momentum valid:

  $\begin{array}{l}{I}_A{\omega }_A=I_o{\omega }_2=C\\ \frac{1}{2}{M}_A{R}^2{\omega }_A=\frac{1}{2}(M_A+M_B)R^2{\omega }_B\\ M_A{\omega }_A=(M_A+M_B){\omega }_B\mathrm{...}(2)\end{array}$

Here, C is the constant.

Where, $I_A$ is moment of inertia.

  ${\omega }_A,{\omega }_B$ is angular velocity.

  $M_A,M_B$ is mass of disc.

  $R$ is radius of disc.

Calculation:

Substitute the given values in equation (2):

  $\begin{array}{l}6(7.2)=(6+9){\omega }_2\\ {\omega }_2=\frac{43.2}{15}=2.88\text{rad/s}\\ {\omega }_2=2.9\text{rad/s}\end{array}$

Conclusion:

The value of angular velocity is $2.9\text{rad/s}$ .