At some instant, a particle traveling in a horizontal circular path of radius 7.50 m has a total acceleration with a magnitude of 16.0 m/s^2 and a constant tangential acceleration of 12.0 m/s^2. Determine the speed of the particle at this instant and (1/8) revolution later.
Hey sorry! My browser is doing weird things and didn't allow me to see this had already posted! Didn't mean to repeat.
To determine the speed of the particle at the given instant, we can divide the total acceleration into its tangential and centripetal components.
The tangential acceleration, denoted as at, is responsible for changing the speed of the particle. Given that the tangential acceleration is constant, we can find it using the following formula:
at = α * r
where α represents the angular acceleration and r is the radius of the circular path. Since the particle is traveling in a horizontal circular path, α is equal to the rate of change of the particle's speed divided by the radius:
α = a / r
Substituting the given values:
at = (12.0 m/s^2) * (7.50 m)
at = 90 m/s^2
Now, let's find the centripetal acceleration, which is responsible for keeping the particle moving in a circular path. The centripetal acceleration, denoted as ac, can be obtained using the formula:
ac = v^2 / r
where v represents the speed of the particle and r is the radius. Rearranging the equation, we get:
v = √(ac * r)
Substituting the given values:
ac = (16.0 m/s^2) - (12.0 m/s^2)
ac = 4.0 m/s^2
v = √(4.0 m/s^2 * 7.50 m)
v = √30 m^2/s^2
v = 5.48 m/s (approximately)
Therefore, the speed of the particle at the given instant is approximately 5.48 m/s.
To determine the speed of the particle at (1/8) revolution later, we need to consider that the particle experiences a tangential acceleration of 12.0 m/s^2. We can calculate the change in speed using the equation:
Δv = at * t
where Δv is the change in speed, at is the tangential acceleration, and t is the time. As the particle completes (1/8) revolution, the time it takes can be calculated using the equation:
t = (1/8) * T
where T represents the period of the circular motion.
The period of the motion can be found using the formula:
T = 2π * r / v
where r is the radius of the circular path and v is the initial speed of the particle. However, since we are trying to find the speed (v') after (1/8) revolution, we need to solve for v' instead of v. Rearranging the equation, we get:
v' = 2π * r / T
Plugging in the values:
v' = 2π * (7.50 m) / T
To find T, we divide the circumference (2π * r) by the initial speed (v):
T = (2π * r) / v
Substituting the given values:
T = (2π * (7.50 m)) / (5.48 m/s)
T ≈ 2.72 s
Now that we know the period, we can calculate the time it takes for (1/8) revolution:
t = (1/8) * T
t ≈ (1/8) * 2.72 s
t ≈ 0.34 s
Using the equation for the change in speed:
Δv = at * t
Δv = (12.0 m/s^2) * (0.34 s)
Δv ≈ 4.08 m/s (approximately)
Therefore, the speed of the particle at (1/8) revolution later is approximately 5.48 m/s + 4.08 m/s = 9.56 m/s.