A race car, starting from rest, travels around a circular turn of radius 29.3 m. At a certain instant, the car is still accelerating, and its angular speed is 0.621 rad/s. At this time, the total acceleration (centripetal plus tangential) makes an angle of 35.0° with respect to the radius. (The situation is similar to that in Figure 8.15b.) What is the magnitude of the total acceleration?
Well, well, well, we've got a speeding race car on our hands! Let's unravel this racing mystery, shall we?
First things first, we need to find the centripetal acceleration, which is responsible for keeping the car on a circular path. We can calculate it using the formula:
ac = ω²r
Where ω is the angular speed and r is the radius. Plugging in the values we have, we get:
ac = (0.621 rad/s)² * 29.3 m
Now, let's move on to the tangential acceleration. The total acceleration is the vector sum of the centripetal and tangential accelerations. We are given that the angle between the total acceleration and the radius is 35.0°.
But here's the twist, my friend: tangential acceleration acts perpendicular to the radius. So, if we draw a right triangle with the radius as the hypotenuse, the total acceleration would be the hypotenuse, the centripetal acceleration would be the adjacent side, and the tangential acceleration would be the opposite side.
Using some trigonometry, we can find the magnitude of the total acceleration. We have:
cos(35.0°) = adjacent / hypotenuse
cos(35.0°) = ac / total acceleration
Solving for the total acceleration, we get:
total acceleration = ac / cos(35.0°)
total acceleration = [ (0.621 rad/s)² * 29.3 m ] / cos(35.0°)
Now, we just need to grab our trusty calculator, crunch some numbers, and voila! We'll have the magnitude of the total acceleration.
Remember, my friend, stay safe on the race track and keep those wheels spinning!
To find the magnitude of the total acceleration, we need to find the centripetal acceleration and the tangential acceleration and then calculate their vector sum.
Given:
Radius (r) = 29.3 m
Angular speed (ω) = 0.621 rad/s
Angle (θ) = 35.0°
First, let's find the centripetal acceleration (a_c), which is given by the formula:
a_c = r * ω^2
Substituting the given values:
a_c = (29.3 m) * (0.621 rad/s)^2
Calculating:
a_c = 11.584 m/s^2 (rounded to three decimal places)
Next, let's find the tangential acceleration (a_t), which is the acceleration component parallel to the tangent of the circular path. This can be found using the formula:
a_t = r * α
Where α is the angular acceleration. Since the problem doesn't provide the value of α, we can assume that the car is starting from rest. In this case, α = ω^2 / r
Substituting the given values:
α = (0.621 rad/s)^2 / 29.3 m
Calculating:
α = 0.013115 m/s^2 (rounded to six decimal places)
Now, we have both the centripetal acceleration and the tangential acceleration. The total acceleration (a_total) is the vector sum of these two accelerations. Using the given angle (θ), we can find the magnitude of the total acceleration using the formula:
a_total = √(a_c^2 + a_t^2 + 2 * a_c * a_t * cosθ)
Substituting the values:
a_total = √((11.584 m/s^2)^2 + (0.013115 m/s^2)^2 + 2 * 11.584 m/s^2 * 0.013115 m/s^2 * cos(35.0°))
Calculating:
a_total = 11.694 m/s^2 (rounded to three decimal places)
Therefore, the magnitude of the total acceleration is 11.694 m/s^2.
To find the magnitude of the total acceleration, we can break it down into its radial and tangential components. The radial component is responsible for the centripetal acceleration, while the tangential component is responsible for the change in speed of the car.
Given:
- Radius of the circular turn (r) = 29.3 m
- Angular speed of the car (ω) = 0.621 rad/s
- Angle between the total acceleration and the radius (θ) = 35.0°
First, let's find the centripetal acceleration using the formula:
ac = rω²
Substitute the given values:
ac = (29.3 m)(0.621 rad/s)²
Calculate the value of ac:
ac ≈ 11.114 m/s²
Next, let's find the tangential acceleration. We know that the total acceleration makes an angle of 35.0° with respect to the radius. We can use trigonometry to find the tangential acceleration.
The tangential acceleration (at) can be found using the formula:
at = ac * tan(θ)
Substitute the known values:
at = 11.114 m/s² * tan(35.0°)
Calculate the value of at:
at ≈ 6.191 m/s²
Now, we have both the centripetal acceleration (ac) and the tangential acceleration (at). To find the magnitude of the total acceleration (a), we can use the Pythagorean theorem:
a = √(ac² + at²)
Substitute the known values:
a = √((11.114 m/s²)² + (6.191 m/s²)²)
Calculate the value of a:
a ≈ √(123.610996 + 38.277481)
a ≈ √161.888477
a ≈ 12.733 m/s²
Therefore, the magnitude of the total acceleration is approximately 12.733 m/s².
The centripetal part of the acceleration is
a,c = r w^2 = (29.3)*(0.621)^2 = 11.32 m/s^2
It is directed toward the center of the circle. The tangential acceleration a,t is such that
a,t/a,c = tan 35 = 0.700
The tangential acceleration is therefore
a,t = 16.17 m/s^2
The total acceeration a is such that
(a,c)/ a = sin 35
a = 19.74 m/s^2