a wheel rotating at 1800 rpm slows down uniformly to 1200 rpm in 2.0 sec. calculate a)the angular acceleration (revs/s^2) of the motor. and b) the number of revolutions it makes in this time.
number of revolutions:
d=avgspeed*time=1500rpm*2/60 min
acceleration:
wf=wi+at
a= (wf-wi)/t=600rpm*2PIrad/rev*1m/60sec*1/2sec
Good
To calculate the angular acceleration (revs/s²) of the motor, you can use the formula:
Angular acceleration (α) = (Final angular velocity - Initial angular velocity) / Time
Let's calculate it:
a) The initial angular velocity is given as 1800 rpm, which is equal to (1800 rev/min) * (1 min/60 sec) = 30 rev/sec.
The final angular velocity is given as 1200 rpm, which is equal to (1200 rev/min) * (1 min/60 sec) = 20 rev/sec.
The time taken is given as 2.0 seconds.
Using the formula, we can plug in the values:
Angular acceleration (α) = (20 rev/sec - 30 rev/sec) / 2.0 sec
= (-10 rev/sec) / 2.0 sec
= -5 rev/sec²
Therefore, the angular acceleration of the motor is -5 rev/sec².
b) To calculate the number of revolutions the wheel makes in 2.0 seconds, we can use the formula:
Number of revolutions = (Initial angular velocity + Final angular velocity) * (Time / 2)
Plugging in the given values:
Number of revolutions = (30 rev/sec + 20 rev/sec) * (2.0 sec / 2)
= 50 rev/sec * 1.0 sec
= 50 revolutions
Therefore, the wheel makes 50 revolutions in 2.0 seconds.