A circular disc is rotating about its own axis at constant angular acceleration. If its angular velocity increases from 210 rpm to 420 rpm during 21 rotations then the angular acceleration of disc is

Nikhal laude, pehile pursath me nikhal

To find the angular acceleration of the disc, we can use the formula:

Angular acceleration (α) = (change in angular velocity) / (time taken)

First, let's convert the given angular velocities from rpm (revolutions per minute) to radians per second, as the SI unit for angular velocity is radians per second.

The conversion factor is:
1 revolution = 2π radians

Angular velocity at the initial state:
ω1 = 210 rpm
= 210 * 2π radians/minute
= 210 * 2π / 60 radians/second

Angular velocity at the final state:
ω2 = 420 rpm
= 420 * 2π radians/minute
= 420 * 2π / 60 radians/second

Now, let's calculate the change in angular velocity:
Δω = ω2 - ω1

Next, we'll calculate the time taken for the change in angular velocity. The problem states that the disc completes 21 rotations during this time.

Since one rotation is equivalent to 2π radians, we can calculate the total angular displacement during this time period:

Total angular displacement = 21 rotations * 2π radians/rotation

Finally, we can calculate the time taken using the formula:

Time taken = Total angular displacement / ω1

Now we have all the necessary values to calculate the angular acceleration:

Angular acceleration (α) = Δω / Time taken

By substituting the values into the formula, we can find the angular acceleration of the disc.

final=initial +a*time

average speed=(420+210)/2=315rpm
timefor 21 rotations=21*60/315=4 sec
a= (final-initial)/time= now we need to get units straight, so lets go to rev/sec, and time in sec
a=(420-210)/60 /4 rev/sec^2=210/240=7/8 rev /sec^2

check my math.

210rpm = 210/60 = 3.5 rev/s.

420rpm = 420/60 = 7.0 rev/s.

T1 = 21/3.5 = 6 s.
T2 = 21/7 = 3 s.

V = Vo + a*(T1-T2).
7 = 3.5 + a*(6-3),
a = 1.17 rev/s^2.

7 = 3.5 + a*(6-3).
a = 1.17 rev/s^2.

I want now the clear solution