Given:
initial angular speed, = 21.5 rad/s
final angular speed, = 28.0 rad/s
time, t = 3.50 s
Solution:
Angular acceleration can be defined as the time rate of change of angular velocity and is given by:
Now, putting the given values in the above formula:
Therefore, angular acceleration is:
Answer:
a) -2.1731 m⁻s
Explanation:
L= length of thread, H= height of ball from ceiling,
N/B h= l cosФ, Ф=31°, l=2.7m, g=9.8m⁻s, m= 6kg
Required to find speed v, of the ball in m⁻s
ω=√g/h, =√g/lcosФ
⇒ ω=√9.8/2.7×cos31°
= 1.992rad⁻s
Now, speed in m/s, v= r.ω
where tanФ= r/h ⇒
r= h×cos 31°
r= 2.7 × cos(31) × tan(31) =-1.0909m
v= 1.992× -1.0909 = -2.1731 m/s
