An automobile travelling with an initial velocity of 25m/s is accelerated to 35m/s in 6 second. The wheel of automobile is 80cm in diameter. Find the angular acceleration
radius= .4m
wi=25/.4 wf= 35/.4
angacceleration= (wf-wi)/time
From the Eqn of motion we have: v=u+at
=25+35*6
=360m/s
recall that angacceleration=v^2/r
r=40m,
=360^2*40
=3240m/s^2
To find the angular acceleration of the wheel, we first need to convert the linear velocity to angular velocity.
The linear velocity of a point on the outer circumference of a wheel is given by the formula:
v = ω * r
where v is the linear velocity, ω (omega) is the angular velocity, and r is the radius of the wheel.
In this case, the wheel diameter is given as 80 cm, which means the radius is half of that, or 40 cm. To convert this to meters, we divide by 100:
r = 40 cm / 100 = 0.4 m
The initial and final linear velocities of the car are given as 25 m/s and 35 m/s, respectively.
For the initial velocity:
25 = ω * 0.4
ω = 25 / 0.4
ω = 62.5 rad/s
For the final velocity:
35 = ω * 0.4
ω = 35 / 0.4
ω = 87.5 rad/s
Now that we have the initial and final angular velocities, we can calculate the angular acceleration using the formula:
α = (ωf - ωi) / t
where α is the angular acceleration, ωf is the final angular velocity, ωi is the initial angular velocity, and t is the time.
In this case, the time given is 6 seconds.
Substituting the values:
α = (87.5 - 62.5) / 6
α = 25 / 6
α ≈ 4.17 rad/s²
Therefore, the angular acceleration of the wheel is approximately 4.17 rad/s².