a wheel starts from rest on application of torque which gives it an angular acceleration @=2t^2-t^2 for first two seconds after which @= 0 then find angular velocity of wheel after 4 seconds.
To find the angular velocity of the wheel after 4 seconds, we need to calculate the change in angular velocity during the time interval of 0 to 4 seconds.
Given:
Angular acceleration, α = 2t^2 - t^2
Where t is the time in seconds.
First, let's find the angular acceleration for the first two seconds (0 to 2 seconds). We substitute t = 2 into the equation:
α = 2(2^2) - 2^2
= 8 - 4
= 4 rad/s^2
Next, we need to find the angular acceleration for the time from 2 to 4 seconds. Since α becomes zero after 2 seconds, the angular acceleration during this interval is zero.
Now, we can calculate the change in angular velocity during each time interval.
For the first two seconds:
Using the equation of motion for angular velocity, we have:
ω = ω0 + αt
= 0 + 4 × 2
= 8 rad/s
For the next two seconds (2 to 4 seconds):
Since α = 0, the angular velocity remains constant during this interval, so:
ω = 8 rad/s
Therefore, the angular velocity of the wheel after 4 seconds is 8 rad/s.
so the angular velocity after four seconds is the same after two seconds, as after two seconds, no more acceleration. 2t^2-t^2 is an odd algebra function, as it reduces quickly to t^2
I don't know your math level, calculus or not.
If you have not had calculus, do this.
Plot angular acceleration vs time from zero to t=2. Now compute the area under that plot: the area is angular velocity (acceleration vs time=velocity). So use a graph paper which lets you count squares. Some graphing calulators will measure the area under the graph for you.
If you are in integral calculus:
velocity=INTegral a(t)dt from zero to 2
= INT (2t^2-t^2)dt=t^3/3 fromzero to 2
= 8/4 rad /sec
Check you acceleration function to verify it is 2t^2-t^2. Even in my Texas, that is odd.