Question

A potter's wheel is spinning with an initial angular velocity of 11 rad/s . It rotates through an angle of 80.0 rad in the process of coming to rest.
What was the angular acceleration of the wheel?
How long does it take for it to come to rest?

Answer 1

• The angular acceleration of the wheel approximately -0.76 rad/s² or proportionally as deceleration approximately 0.76 rad/s.
• It need approximately 14.474 s to come to rest.

Introduction

Hi ! I will help you to discuss about Proportionally Changes in Circular Motion. The analogy of proportionally changes in circular motion is same as the analogy of proportionally changes in direct motion. Here you will hear again the terms acceleration and change in speed, only expressed in the form of a certain angle coverage. Before that, in circular motion, it is necessary to know the following conditions:

• 1 rotation = 2π rad
• 1 rps = 2π rad/s
• 1 rpm = $\sf{\frac{1}{60} \: rps}$ = $\sf{\frac{1}{30}\pi \: rad/s}$

Formula Used

The following equations apply to proportionally changes circular motion:

Relationship between Angular Acceleration and Change of Angular Velocity

$\boxed{\sf{\bold{\omega_t = \omega_0 + \alpha \times t}}}$

With the following conditions:

• $\sf{\omega_t}$ = final angular velocity  (rad/s)
• $\sf{\omega_0}$ =  initial angular velocity (rad/s)
• $\sf{\alpha}$ = angular acceleration (rad/s²)
• t = interval of the time (s)

Relationship between Angular Acceleration and Change of $\sf{\theta}$ (Angle of Rotation)

$\boxed{\sf{\bold{\theta = \omega_0 \times t + \frac{1}{2} \times \alpha \times t^2}}}$

Or

$\boxed{\sf{\bold{(\omega_t)^2= (\omega_0)^2 + 2 \times \alpha \times \theta}}}$

With the following condition :

• $\sf{\theta}$ = change of the sudut (rad)
• $\sf{\alpha}$ = angular acceleration (rad/s²)
• t = interval of the time (s)
• $\sf{\omega_t}$ = final angular velocity  (rad/s)
• $\sf{\omega_0}$ =  initial angular velocity (rad/s)

Problem Solving

We know that :

• $\sf{\omega_t}$ = final angular velocity  = 0 rad/s >> see in the sentence "in the process of coming to rest."
• $\sf{\omega_0}$ =  initial angular velocity = 11 rad/s
• $\sf{\theta}$ = change of the sudut = 80.0 rad

What was asked :

• $\sf{\alpha}$ = angular acceleration = ... rad/s²
• t = interval of the time = ... s

Step by step :

• $\sf{\alpha}$ = ... rad/s²

$\sf{(\omega_t)^2= (\omega_0)^2 + 2 \times \alpha \times \theta}$

$\sf{0^2= (11)^2 + 2 \times \alpha \times 80}$

$\sf{0 = 121 + 160 \alpha}$

$\sf{-160 \alpha = 121}$

$\sf{\alpha = \frac{121}{-160}}$

$\sf{\alpha = -0.75625 \: rad/s^2 \approx \boxed{-0.76 \: rad/s^2}}$

• t = ... s

$\sf{\alpha = \frac{\omega_0 - \omega_t}{t}}$

$\sf{-0.76 = \frac{0 - 11}{t}}$

$\sf{-0.76t = -11}$

$\sf{t = \frac{- 11}{-0.76}}$

$\boxed{\sf{t \approx 14.474 \: s}}$

Conclusion

So :

• The angular acceleration of the wheel approximately -0.76 rad/s² or proportionally as deceleration approximately 0.76 rad/s.
• It need approximately 14.474 s to come to rest.