A jogger is running around a circular track of circumference 340 m. If the jogger has a speed of 11 km/h, what is the magnitude of the centripetal acceleration of the jogger?
A person drives a car around a circular cloverleaf with a radius of 63 m at a uniform speed of 10 m/s.
(a) What is the acceleration of the car?
To find the magnitude of the centripetal acceleration, we first need to calculate the angular velocity of the jogger.
The formula for angular velocity (ω) is given by:
ω = v / r
where v is the linear velocity (in this case, 11 km/h) and r is the radius of the circular track.
We can find the radius by using the formula for circumference:
C = 2πr
Given that the circumference of the track is 340 m, we can solve for the radius:
340 = 2πr
Dividing both sides of the equation by 2π, we get:
r = 340 / (2π)
Now we have the radius, we can calculate the angular velocity:
ω = (11 km/h) / (340 / (2π))
Converting 11 km/h to m/s:
(11 km/h) × (1000 m/km) / (60 s/min) / (60 min/h) = 3.055 m/s
Substituting this value into the formula for angular velocity:
ω = (3.055 m/s) / (340 / (2π))
Now we can calculate the angular velocity.
ω = (3.055 m/s) / (340 / (2π)) = 0.029 rad/s (rounded to three decimal places)
The magnitude of the centripetal acceleration (a) can be found using the formula:
a = ω²r
Substituting the values we have:
a = (0.029 rad/s)² × (340 / (2π))
Simplifying further:
a = (0.000841 m²/s²) × (340 / (2π))
a = 0.034 m/s² (rounded to three decimal places)
Therefore, the magnitude of the centripetal acceleration of the jogger is approximately 0.034 m/s².