A train is rounding a circular curve whose radius is 1.87E+2 m. At one instant, the train has an angular acceleration of 1.51E-3 rad/s2 and an angular speed of 0.0470 rad/s. Find the magnitude of the total acceleration (centripetal plus tangential) of the train.
centripetal acceleration= w^2 r
tangential accceleration= 1.51E-3*r
magnitude= sqrt((w^2r)^2+(1.51E-3r)^2)
= r*sqrt(w^4+2.28E-6)
= r*E-3*sqrt(4.88+2.28)
= .00267 r =0.50 m/s^2
check all that.
a(centripetal) = ω^2•R=(0.047)^2•187=0.413 m/s^2.
a(tangential)=ε•R=1.51•10^-3•187=0.282 m/s^2.
a(total) =sqr(a(centripetal)^2+ a(tangential)^2) =0.5 m/s^2.
Why did the train become a circus performer? Because it wanted to show off its impressive total acceleration! To find the magnitude, we need to add the centripetal acceleration and tangential acceleration.
First, let's find the centripetal acceleration:
The formula for centripetal acceleration is given by a = r * ω^2, where r is the radius and ω is the angular speed.
Plugging in the values, we get a = 1.87E+2 m * (0.0470 rad/s)^2.
Next, let's find the tangential acceleration:
The formula for tangential acceleration is given by a = r * α, where α is the angular acceleration.
Plugging in the values, we get a = 1.87E+2 m * 1.51E-3 rad/s^2.
Now, let's add the two accelerations together to find the total acceleration:
total acceleration = centripetal acceleration + tangential acceleration.
After doing the math, we'll have the magnitude of the total acceleration! Sorry about the math jokes, sometimes my calculations get a little clownfused.
To find the magnitude of the total acceleration of the train, we can separate the centripetal and tangential accelerations and then calculate the total acceleration by using the Pythagorean theorem.
Given:
Radius of the curve, r = 1.87E+2 m
Angular acceleration, α = 1.51E-3 rad/s^2
Angular speed, ω = 0.0470 rad/s
Centripetal Acceleration (ac):
The centripetal acceleration of an object moving in a circular path is given by the formula:
ac = r * ω^2
Substituting the given values:
ac = (1.87E+2 m) * (0.0470 rad/s)^2
Now, let's calculate it:
ac = (1.87E+2) * (0.0470^2) = 0.419 m/s^2
Tangential Acceleration (at):
The tangential acceleration of an object moving in a circular path is given by the formula:
at = r * α
Substituting the given values:
at = (1.87E+2 m) * (1.51E-3 rad/s^2)
Now, let's calculate it:
at = (1.87E+2) * (1.51E-3) = 0.2827 m/s^2
Total Acceleration (a):
The total acceleration can be determined using the Pythagorean theorem:
a = √(ac^2 + at^2)
Substituting the values we calculated earlier:
a = √(0.419^2 + 0.2827^2)
Now, let's calculate it:
a = √(0.175561 + 0.07995529)
a = √0.25551629
a = 0.5055 m/s^2 (rounded to four decimal places)
Therefore, the magnitude of the total acceleration of the train is 0.5055 m/s^2.
To find the magnitude of the total acceleration of the train, we need to consider both the centripetal acceleration and the tangential acceleration.
The centripetal acceleration is the acceleration towards the center of the circular path and is given by the formula:
ac = R * ω^2
where ac is the centripetal acceleration, R is the radius of the curve, and ω is the angular speed.
Substituting the given values, we have:
ac = (1.87E+2 m) * (0.0470 rad/s)^2
Calculating this expression gives us the value of the centripetal acceleration.
Next, we need to find the tangential acceleration. The tangential acceleration is the rate at which the angular speed is changing and is given by the formula:
at = R * α
where at is the tangential acceleration, R is the radius of the curve, and α is the angular acceleration.
Substituting the given values, we have:
at = (1.87E+2 m) * (1.51E-3 rad/s^2)
Calculating this expression gives us the value of the tangential acceleration.
Finally, to find the total acceleration, we need to find the vector sum (magnitude) of the centripetal acceleration and the tangential acceleration:
a = √(ac^2 + at^2)
Plug in the values of ac and at that we calculated earlier, and evaluate the expression to find the magnitude of the total acceleration of the train.