(1500 rev/min)(min / 60 s) / (3.0 s) = 8.33 rev/s²
<span>(B) </span>
<span>(1/2)(8.33 rev/s²)(3.0 s)² = 37.5 rev </span>
<span>(C) </span>
<span>(1500 rev/min)(min / 60 s)[2π(0.12 m) / rev] = 18.8 m/s</span>
Answer:
The angular velocity is
5.64rad/s
Explanation:
This problem bothers on curvilinear motion
The angular velocity is defined as the rate of change of angular displacement it is expressed in rad/s
We know that the velocity v is given as
v= ωr
Where ω is the angular velocity
r is 300mm to meter = 0.3m
the radius of the circle
described by the level
v=1.64m/s
Making ω subject of the formula and solving we have
ω=v/r
ω=1.64/0.3
ω=5.46 rad/s
Answer:
Angular frequency will increase
No change in the amplitude
Explanation:
At extreme end of the SHM the energy of the SHM is given by

here we know that

now at the extreme end when one of the mass is removed from it
then in that case the angular frequency will change

So angular frequency will increase
but the position of extreme end will not change as it is given here that the top block is removed without disturbing the lower block
so here no change in the amplitude
Am sorry what can you be more specific