- #1
Flucky
- 95
- 1
Hi all,
The motor driving a large grindstone is switched off when a rotational speed of 360 rpm has been achieved. After 15s the speed has decreased to 210rpm. If the angular deceleration remains constant, how many additional revolutions does the stone make before coming to rest?
So to start with I converted rpm to rads-1 which was easy enough (I left it in pi to make life easier)
ω0 = 360/60 x 2∏ =12∏ rads-1
ω15 = 210/60 x 2∏ =7∏ rads-1
I then used v=u+at to find the acceleration
7∏ = 12∏ + 15a
a = -∏/3 rads-2
This is where I'm having a mind block as to what to do next, from 15s to when the stone comes to rest - how many additional revolutions does the stone make?
Would love some pointers (I'm sure it's a simple solution but my heads refusing to grasp it).
Homework Statement
The motor driving a large grindstone is switched off when a rotational speed of 360 rpm has been achieved. After 15s the speed has decreased to 210rpm. If the angular deceleration remains constant, how many additional revolutions does the stone make before coming to rest?
The Attempt at a Solution
So to start with I converted rpm to rads-1 which was easy enough (I left it in pi to make life easier)
ω0 = 360/60 x 2∏ =12∏ rads-1
ω15 = 210/60 x 2∏ =7∏ rads-1
I then used v=u+at to find the acceleration
7∏ = 12∏ + 15a
a = -∏/3 rads-2
This is where I'm having a mind block as to what to do next, from 15s to when the stone comes to rest - how many additional revolutions does the stone make?
Would love some pointers (I'm sure it's a simple solution but my heads refusing to grasp it).