Question

A disk, with a radius of 0.25 m, is to be rotated like a merry-go-round through 800 rad, starting from rest, gaining angular speed at the constant rate through the first 400 rad and then losing angular speed at the constant rate - until it is again at rest. The magnitude of the centripetal acceleration of any portion of the disk is not to exceed 100 m/s2. (a) What is the least time required for the rotation

Answer 1

Answer:

a) $T_{min} = 80\,s$

Explanation:

a) Let consider that disk accelerates and decelerates at constant rate. The expression for angular acceleration and deceleration are, respectively:

Acceleration

$\alpha_{1} = \frac{\omega_{max}^{2}}{2\cdot (400\,rad)}$

Deceleration

$\alpha_{2} = -\frac{\omega_{max}^{2}}{2\cdot (400\,rad)}$

Since angular acceleration and deceleration have same magnitude but opposite sign. Let is find the maximum allowed angular speed from maximum allowed centripetal acceleration:

$a_{r,max} = \omega_{max}^{2}\cdot r$

$\omega_{max} = \sqrt{\frac{a_{r,max}}{r} }$

$\omega_{max} = \sqrt{\frac{100\,\frac{m}{s^{2}} }{0.25\,m} }$

$\omega_{max} = 20\,\frac{rad}{s}$

Maximum magnitude of acceleration/deceleration is:

$\alpha = 0.5\,\frac{rad}{s^{2}}$

The least time require for rotation is:

$T_{min} = 2\cdot \left(\frac{20\,\frac{rad}{s} }{0.5\,\frac{rad}{s^{2}} } \right)$

$T_{min} = 80\,s$