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

a)
`T_(min) = 80\,s`

Step-by-step explanation:

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

Acceleration

`\alpha_(1) = (\omega_(max)^(2))/(2\cdot (400\,rad))`

Deceleration

`\alpha_(2) = -(\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{(a_(r,max))/(r) }`

`\omega_(max) = \sqrt{(100\,(m)/(s^(2)) )/(0.25\,m) }`

`\omega_(max) = 20\,(rad)/(s)`

Maximum magnitude of acceleration/deceleration is:

`\alpha = 0.5\,(rad)/(s^(2))`

The least time require for rotation is:

`T_(min) = 2\cdot \left((20\,(rad)/(s) )/(0.5\,(rad)/(s^(2)) ) \right)`

`T_(min) = 80\,s`