A ball starts from rest at point A and rolls down a ramp with uniform angular acceleration, taking 17.5 s to reach the bottom of the ramp. The slope of the ramp is now changed so that the ball has twice the angular acceleration as before. If the ball is again released from rest at point A, the time it will take it to reach the bottom of the ramp is closest to:
a.24.7 s
b.8.75 s
c.13.9 s
d.12.4 s
e.35.0 s
I know the answer is 12.4. I need to know how to do it.
Thank you!
a.24.7 s
b.8.75 s
c.13.9 s
d.12.4 s
e.35.0 s
I know the answer is 12.4. I need to know how to do it.
Thank you!

The ball is rolling it implies it will follow the equation a = r α
In first case let angular acceleration be α, so linear acceleration a = r α
let length of ramp is s and it is given that initial velocity ( u ) = 0 m/sec and time taken ( t ) is 17.5 sec
now use equation of motion, s = ut + ½ at²
s = ½ r α (17.5 )² .................................equatio… 1
In second case angular acceleration is 2α so linear acceleration will be 2rα
let time taken to roll down now is ' t '
=> s = ½ 2rα t² ...................................equat… 2
divide equation 1 by equation 2
you will get t² = (17.5 )²/ 2
so t = 17.5 / √2
= 12.374
In first case let angular acceleration be α, so linear acceleration a = r α
let length of ramp is s and it is given that initial velocity ( u ) = 0 m/sec and time taken ( t ) is 17.5 sec
now use equation of motion, s = ut + ½ at²
s = ½ r α (17.5 )² .................................equatio… 1
In second case angular acceleration is 2α so linear acceleration will be 2rα
let time taken to roll down now is ' t '
=> s = ½ 2rα t² ...................................equat… 2
divide equation 1 by equation 2
you will get t² = (17.5 )²/ 2
so t = 17.5 / √2
= 12.374