A .7m diameter wheel accelerates uniformly from 160rpm to 280rpm in 4.0s.
Determine:
a)it's angular acceleration, and
b)the radial and tangential components of the linear acceleration of a point on the edge of the wheel 2.0s after it has started accelerating.
I want to use the equation t= w2-w1/a for part a, and when I did I got a= 30, but this doesn't seem right. Is there a better way to do this problem?
To determine the angular acceleration of the wheel, you correctly utilized the equation:
t = (w2 - w1) / a
where t is time, w2 is the final angular velocity, w1 is the initial angular velocity, and a is the angular acceleration.
Given that w2 = 280 rpm, w1 = 160 rpm, and t = 4.0 seconds, you can now determine the angular acceleration:
a = (w2 - w1) / t
= (280 rpm - 160 rpm) / 4.0 s
= 120 rpm / 4.0 s
= 30 rpm/s
However, it's worth noting that the angular acceleration is usually measured in radians per second squared (rad/s²) rather than rpm/s. So we need to convert rpm/s to rad/s²:
1 revolution = 2π radians
1 minute = 60 seconds
Thus, 1 rpm = (2π radians) / (60 seconds)
Therefore, you can convert the angular acceleration from rpm/s to rad/s²:
a = (30 rpm/s) * (2π rad / 60 s)
≈ 3.14 rad/s²
So, after correcting the unit and rounding to two decimal places, the angular acceleration of the wheel is approximately 3.14 rad/s².
For part b, to determine the radial and tangential components of the linear acceleration at a point on the edge of the wheel 2.0 s after it starts accelerating, you need the formula:
a = r * α
where a is the linear acceleration, r is the radius of the wheel, and α is the angular acceleration.
Given that the diameter of the wheel is 0.7 m, the radius is half of that, r = 0.7 m / 2 = 0.35 m.
Using the angular acceleration from part a, α = 3.14 rad/s², you can now calculate the linear acceleration at the edge of the wheel:
a = (0.35 m) * (3.14 rad/s²)
≈ 1.1 m/s²
So, after performing the calculation, the linear acceleration at a point on the edge of the wheel, 2.0 s after it starts accelerating, is approximately 1.1 m/s².
To find the radial component of the linear acceleration, simply multiply the linear acceleration by the sine of the angle (θ) formed by the radius vector with the horizontal line:
aradial = a * sin(θ)
Since the radius vector is horizontal, the angle θ is 0°. Therefore, sin(0°) = 0.
Thus, the radial component of the linear acceleration is aradial = 0 m/s².
To find the tangential component of the linear acceleration, multiply the linear acceleration by the cosine of the angle (θ) formed by the radius vector with the horizontal line:
atangential = a * cos(θ)
Since the radius vector is horizontal, the angle θ is 0°. Therefore, cos(0°) = 1.
Thus, the tangential component of the linear acceleration is
atangential = a * cos(θ)
= (1.1 m/s²) * (cos(0°))
= 1.1 m/s²
So, the tangential component of the linear acceleration is atangential = 1.1 m/s².
In summary:
a) The angular acceleration of the wheel is approximately 3.14 rad/s².
b) The radial component of the linear acceleration is 0 m/s², and the tangential component of the linear acceleration is 1.1 m/s².