Chapter 8, Problem 30P

(a)

To determine

To find: The torque being larger in which the diameter is $12\text{cm}$ .

Answer to Problem 30P

The friction force between her hands and clay is $9\times {10}^{-2}\text{N}\cdot \text{m}$ .

Explanation of Solution

Given information:

A potter is shaping a bowl on a potter’s wheel rotating at constant angular speed. The friction force between her hands and the clay is $1.5\text{N}$ totals.

Formula used:

The torque on the wheel

  $\tau =F_r\times \text{radius}$

Here,

  $\tau$ is the torque,

  $F_r$ is the friction force between hands and clay.

Calculation:

Friction force between her hands and clay $F_r=1.5\text{N}$

Diameter of bowl $\left(D\right)=0.12\text{m}$

The torque on the wheel $\tau =F_r\times \text{radius}$

Substituting the value as shown below,

  $\begin{array}{l}\tau =F_r\times \text{radius}\\ \text{=1}\text{.5}\times \text{0}\text{.06}\\ \tau \text{=9}\times {\text{10}}^{-2}\text{N}\cdot \text{m}\end{array}$

Hence, friction force between her hands and clay is $9\times {10}^{-2}\text{N}\cdot \text{m}$ .

Conclusion:

The friction force between her hands and clay is $9\times {10}^{-2}\text{N}\cdot \text{m}$ .

(b)

To determine

To find: The time that is required for the plotter wheel rotates.

Answer to Problem 30P

The value of time is $t=12.3\text{s}$ .

Explanation of Solution

Given information:

A potter is shaping a bowl on a potter’s wheel rotating at constant angular speed. The friction force between her hands and the clay is $1.5\text{N}$ totals.

Formula used:

The torque $\tau =I\alpha$

Here,

  $\tau$ is the torque.

  $I$ is the moment of inertia.

  $\alpha$ is the angular acceleration.

Calculation:

Initial angular velocity, ${\omega }_0=1.6\text{rev/s}$

Final angular velocity, $\omega =0$

M.I of wheel and bowl, $I=0.11\text{kg}\cdot {\text{m}}^{\text{2}}$

Torque, $\tau =I\alpha$

The net torque is negative, as the object is slowing.

  $\begin{array}{l}-9\times {10}^{-2}=0.11\times \alpha \\ \alpha =-\left(\frac{9\times {10}^{-2}}{0.11}\right){\text{rad/s}}^{\text{2}}\end{array}$

Angular acceleration $\alpha =-0.82{\text{rad/s}}^{\text{2}}$

But the angular acceleration $\alpha =\frac{\omega -{\omega }_0}{t}$

  $t=\frac{0-\left(1.6\times 2\pi \right)}{-0.82}\text{s}$

  $t=12.\text{3}\text{s}$

Conclusion:

The value of time is $t=12.3\text{s}$ .